An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass . The piston and the cylinder have equal cross-sectional area . When the piston is in equilibrium, the volume of the gas is and its pressure is . The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
1
2
3
4
Official Solution
Correct Option: (3)
In equilibrium, ...(i) when slightly displaced downwards.
Restoring force,
Therefore, motion is simple harmonic comparing with F = - kx we have
02
PYQ 2014
medium
physicsID: jee-main
A particle moves with simple harmonic motion in a straight
line. In first sec, after starting from rest it travels a distance a and in next sec, it travels , in same direction, then
1
amplitude of motion is 3a
2
time period of oscillations is
3
amplitude of motion is
4
time period of oscillations is
Official Solution
Correct Option: (4)
As it starts from rest, we have At when when
As
Now,
03
PYQ 2014
easy
physicsID: jee-main
The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from to in seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is , the time in which amplitude of this pendulum will reduce from to carbon dioxide will be close to ( .
1
231 s
2
208 s
3
161 s
4
142 s
Official Solution
Correct Option: (3)
In
In
04
PYQ 2015
medium
physicsID: jee-main
A pendulum made of a uniform wire of cross sectional area has time period . When an additional mass is added to its bob, the time period changes to . If the Young's modulus of the material of the wire is then is equal to ( gravitational acceleration)
1
2
3
4
Official Solution
Correct Option: (1)
Also:
05
PYQ 2015
hard
physicsID: jee-main
A simple harmonic oscillator of angular frequency is acted upon by an external force . If the oscillator is at rest in its equilibrium position at , its position at later times is proportional to :
1
2
3
4
Official Solution
Correct Option: (4)
At If At
06
PYQ 2016
medium
physicsID: jee-main
A particle performs simple harmonic motion with amplitude . Its speed is trebled at the instant that it is at a distance from equilibrium position. The new amplitude of the motion is :
1
2
3A
3
4
Official Solution
Correct Option: (4)
Total initial energy at potential energy Kinetic energy at If speed is tripled, new Kinetic energy New total energy If next amplitude = A' ; then
07
PYQ 2017
medium
physicsID: jee-main
A particle is executing simple harmonic motion with a time period . At time , it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like :
1
2
3
4
Official Solution
Correct Option: (4)
08
PYQ 2018
medium
physicsID: jee-main
An oscillator of mass is at rest in its equilibrium position in a potential . A particle of mass comes from right with speed u and collides completely inelastically with and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after collisions is :
1
2
3
4
Official Solution
Correct Option: (1)
The correct option is(A):
Potential of the given oscillator is Given: Initial momentum of the particle of mass Momentum of (oscillator + particle) after collision Velocity of oscillator after collision So, momentum of system From conservation of linear momentum, we have
For second collision, oscillator and particle have momentum in opposite direction. Net or total momentum is zero. Likewise after collision the momentum is zero. After collision, Mass of oscillator and 12 particles will be Now, from conservation of linear momentum, for collision, we have
Total mass after collision Kinetic energy of system
09
PYQ 2018
medium
physicsID: jee-main
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of . What is the force constant of the bonds connecting one atom with the other ? ( Mole wt. of silver and Avogadro number )
1
6.4 N/m
2
7.1 N/m
3
2.2 N/m
4
5.5 N/m
Official Solution
Correct Option: (2)
10
PYQ 2019
medium
physicsID: jee-main
A particle undergoing simple harmonic motion has time dependent displacement given by . The ratio of kinetic to potential energy of this particle at will be :
1
2
2
3
3
4
1
Official Solution
Correct Option: (3)
Hence ratio is (most appropriate)
11
PYQ 2019
medium
physicsID: jee-main
A cylindrical plastic bottle of negligible mass is filled with of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency . If the radius of the bottle is then close to : (density of water = )
1
2
3
4
Official Solution
Correct Option: (4)
Extra Boyant force = Axg
12
PYQ 2019
medium
physicsID: jee-main
A rod of mass and length is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of are attached at distance from its centre on both sides, it reduces the oscillation frequency by . The value of ratio is close to :
1
0.17
2
0.37
3
0.57
4
0.77
Official Solution
Correct Option: (2)
Frequency of torsonal oscillations is given by
13
PYQ 2019
medium
physicsID: jee-main
A particle is executing simple harmonic motion (SHM) of amplitude A, along the -axis, about . When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be :
1
2
3
4
A
Official Solution
Correct Option: (3)
Potential energy Kinetic energy According to the question, U = k
14
PYQ 2019
easy
physicsID: jee-main
A body of mass falls freely from a height of on a platform of mass which is mounted on a spring having spring constant . The body sticks to the platform and the spring's maximum compression is found to be . Given that , the value of will be close to :
1
4 cm
2
8 cm
3
80 cm
4
40 cm
Official Solution
Correct Option: (1)
Velocity of block just before it collides with block Applying momentum conversation just before and just after collision.
initial compression of spring
applying work energy theorem,
solving
15
PYQ 2019
medium
physicsID: jee-main
A simple pendulum of length is oscillating with an angular frequency . The support of the pendulum starts oscillating up and down with a small angular frequency of and an amplitude of . The relative change in the angular frequency of the pendulum is best given by :
1
2
3
4
Official Solution
Correct Option: (1)
Angular frequency of pendulum
[ = angular frequency of support] rad/sec.
16
PYQ 2019
medium
physicsID: jee-main
A simple pendulum, made of a string of length and a bob of mass m, is released from a small angle . It strikes a block of mass , kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle . Then M is given by :
1
2
3
4
Official Solution
Correct Option: (4)
By momentum conservation
and ...(I) ....(II) Dividing
By componendo divided
17
PYQ 2021
medium
physicsID: jee-main
A bob of mass 'm' suspended by a thread of length undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density times that of the bob and the length of the thread is increased by of the original length, then the time period of the simple harmonic oscillations will be :
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The time period of a simple pendulum is determined by its effective length and the effective acceleration due to gravity ( ). When immersed in a liquid, the buoyant force reduces the effective weight, thereby reducing . Step 2: Key Formula or Approach: 1. Original Time Period: 2. Effective gravity in liquid: Step 3: Detailed Explanation: Given: - Density of liquid (where is bob density). - New length . First, calculate the effective acceleration due to gravity ( ) in the liquid: Now, calculate the new time period : Since , we get: Step 4: Final Answer: The new time period is .
18
PYQ 2021
medium
physicsID: jee-main
Two simple harmonic motion, are represented by the equations Ratio of amplitude of to . The value of x is _________.
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We are given two equations of SHM. We need to find the amplitude of each motion and then find the ratio of their amplitudes to determine the value of x. Step 2: Key Formula or Approach:
1. The standard form of an SHM equation is , where A is the amplitude.
2. An expression of the form can be converted to , where the amplitude . Step 3: Detailed Explanation: Amplitude of :
The first equation is already in the standard form:
By comparing with , we can see that the amplitude of is . Amplitude of :
The second equation is:
Let's first simplify the expression in the parenthesis: .
This is in the form with , , and .
The amplitude of this part is .
So, for some phase .
Now, substitute this back into the equation for :
.
The amplitude of is . Ratio of Amplitudes:
The ratio of the amplitude of to is:
We are given that this ratio is , which means .
Comparing the two, we get . Step 4: Final Answer:
The value of x is 1.
19
PYQ 2021
medium
physicsID: jee-main
In the given figure, a mass is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is . The mass oscillates on a frictionless surface with the time period and amplitude . When the mass is in an equilibrium position, as shown in the figure, another mass is gently fixed upon it. The new amplitude of oscillation will be :
1
2
3
4
Official Solution
Correct Option: (2)
Let's determine the new amplitude of oscillation after another mass is gently placed on the existing mass at equilibrium.
Initially, the mass attached to a spring with constant oscillates with amplitude . The time period of such a simple harmonic oscillator is given by:
When the mass is at equilibrium, and mass is placed on it without any external force, the mass will continue oscillating but with a total mass of .
The energy in the system initially (when mass alone is oscillating) is:
After adding the extra mass, the energy in the system when at maximum amplitude becomes:
Since the total mechanical energy of the system is conserved (in the absence of external forces and damping):
Therefore, equating the expressions and solving for :
The new amplitude of oscillation is , corresponding to the option:
20
PYQ 2021
medium
physicsID: jee-main
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy U(x) versus time (t) plot of the particle is correctly shown in figure :
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
Step 1: Understanding the Question:
We are given a displacement-time ( ) graph for a particle in Simple Harmonic Motion (SHM). We need to identify the correct potential energy-time ( ) graph for the same motion. Step 2: Key Formula or Approach:
The potential energy ( ) of a particle in SHM is given by , where is the spring constant and is the displacement from the mean position.
The displacement as a function of time for the given graph is .
Substituting into the potential energy formula gives . Step 3: Detailed Explanation:
From the formula , we can deduce the key features of the potential energy graph:
1. Non-negativity: Since is positive and is always non-negative, the potential energy must always be greater than or equal to zero ( ). This eliminates any graph that goes below the time axis (like option A).
2. Value at mean position: At the mean position, the displacement is zero ( ). The given graph shows this at points O, B, etc. At these points, the potential energy must be zero: . This eliminates graphs where the minimum value is greater than zero (like option C).
3. Value at extreme positions: At the extreme positions (amplitude), the displacement is maximum ( ). The given graph shows this at points A (maximum positive displacement) and C (maximum negative displacement). At these points, the potential energy is maximum: .
4. Frequency: The displacement is . The potential energy is . Using the identity , we get . The angular frequency of the potential energy oscillation is , which is double the frequency of the displacement oscillation. This means that for every one full cycle of displacement, the potential energy completes two full cycles. Step 4: Final Answer:
Let's check the options against these criteria:
- Option (A) is incorrect because energy cannot be negative.
- Option (C) is incorrect because potential energy is zero at the mean position.
- Option (B) correctly shows that , at points corresponding to O and B on the graph, and at points corresponding to A and C. It also shows two energy cycles for one displacement cycle.
- Option (D) might have similar features, but Graph (B) is the standard representation matching all criteria. Therefore, the graph with ID 86435168204 is the correct representation.
21
PYQ 2021
medium
physicsID: jee-main
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is :
1
86.4 s
2
8.64 s
3
43.2 s
4
4.32 s
Official Solution
Correct Option: (3)
Step 1: Understanding the Question:
The length of a pendulum clock increases, which will change its time period. We need to find the total error in time (time lost or gained) over a period of one day. Step 2: Key Formula or Approach:
The time period (T) of a simple pendulum is given by .
For small changes, the fractional change in the time period can be found using error analysis:
The total error in time ( ) in a day is this fractional change multiplied by the total seconds in a day. Step 3: Detailed Explanation:
Given that the length (L) increases by 0.1%, the fractional change in length is:
Now, we calculate the fractional change in the time period (T):
Since the length increases, the time period also increases ( is positive). This means the clock runs slower and loses time. The total number of seconds in one day is:
The total error in time per day ( ) is: Step 4: Final Answer:
The error in time per day is 43.2 seconds. The clock will lose 43.2 seconds. This corresponds to option (C).
22
PYQ 2021
medium
physicsID: jee-main
Two simple harmonic motions are represented by the equations and . The amplitude of second motion is __________ times the amplitude in first motion.
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We are given two equations representing simple harmonic motions. We need to find the ratio of the amplitude of the second motion to the amplitude of the first motion. Step 2: Key Formula or Approach:
1. The standard form for a simple harmonic motion is , where A is the amplitude.
2. The first equation is already in this standard form.
3. The second equation needs to be converted to the standard form using the trigonometric identity: , where and . Step 3: Detailed Explanation: First Motion:
The equation is .
Comparing this with the standard form , the amplitude of the first motion is . Second Motion:
The equation is .
Let's simplify the term in the parenthesis: .
Here, and . The resultant amplitude of this part is .
The phase angle is given by , so .
Therefore, .
Substitute this back into the equation for :
Comparing this with the standard form, the amplitude of the second motion is . Ratio of Amplitudes:
The question asks for the ratio . Step 4: Final Answer:
The amplitude of the second motion is 2 times the amplitude of the first motion.
23
PYQ 2021
medium
physicsID: jee-main
If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
1
16 m/s
2
π ms
3
9.8 ms
4
2π ms
Official Solution
Correct Option: (4)
Step 1: The time period of a simple pendulum is .
Step 2: Given s and m.
Step 3: Substitute values: .
Step 4: Squaring both sides: ms .
24
PYQ 2021
medium
physicsID: jee-main
Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is:
1
2
3
4
Official Solution
Correct Option: (3)
When the mass 'm' is displaced by a distance 'x' to the right from its equilibrium position, the left spring is stretched by 'x' and the right spring is compressed by 'x'. The restoring force from the left spring is (pointing left). The restoring force from the right spring is (also pointing left). The total restoring force on the mass is the sum of these two forces: . The total force is of the form , where the effective spring constant is . This arrangement is equivalent to two springs connected in parallel. The time period 'T' of a simple harmonic oscillator is given by the formula . Substituting the value of : .
25
PYQ 2021
medium
physicsID: jee-main
Y = A sin( t+ ) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is Y = and it is moving along negative x-direction. Then the initial phase angle will be:
1
2
3
4
Official Solution
Correct Option: (2)
The displacement equation is given by . At , the displacement is . Substituting into the equation: . This condition is satisfied for and in the range . The velocity of the particle is given by the derivative of the displacement: . At , the particle is moving in the negative direction, which means . . Since A and are positive, this requires . Now we check the two possible values for : If , then , which is positive. This is not the correct phase. If , then , which is negative. This matches the condition. Therefore, the initial phase angle is .
26
PYQ 2021
medium
physicsID: jee-main
For a body executing S.H.M. : (a) Potential energy is always equal to its K.E. (b) Average potential and kinetic energy over any given time interval are always equal. (c) Sum of the kinetic and potential energy at any point of time is constant. (d) Average K.E. in one time period is equal to average potential energy in one time period. Choose the most appropriate option from the options given below :
1
only (b)
2
(b) and (c)
3
only (c)
4
(c) and (d)
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept: Simple Harmonic Motion (SHM) involves a continuous transformation between kinetic energy (K.E.) and potential energy (P.E.), while their sum remains constant in the absence of dissipative forces. Step 2: Detailed Explanation: (a) Potential energy and Kinetic energy . They are equal only when , not always. So (a) is false. (b) Average values are only guaranteed to be equal over a full period or specific symmetric intervals. "Any given interval" is too broad. So (b) is false. (c) Total Energy . This is constant throughout the motion. So (c) is true. (d) Over a full time period , both the average potential energy and average kinetic energy are equal to . So (d) is true. Step 3: Final Answer: Statements (c) and (d) are correct.
27
PYQ 2021
medium
physicsID: jee-main
A particle of mass is hanging from a spring of force constant . The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is . The value of x is .
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: In Simple Harmonic Motion (SHM), the total mechanical energy is conserved and alternates between Kinetic Energy (KE) and Potential Energy (PE). We need to find the specific time relative to the period when . Step 2: Key Formula or Approach: 1. Displacement equation: . 2. Potential Energy: . 3. Kinetic Energy: . Step 3: Detailed Explanation: Given the condition:
Substitute this into the displacement equation (assuming the particle starts from the mean position at ):
The first time this occurs is at:
Since :
Comparing this with the given form , we find . Step 4: Final Answer: The value of x is 8.
28
PYQ 2021
medium
physicsID: jee-main
In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ________ rad/s when .
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: For two masses connected by a spring on a frictionless surface, the system oscillates about its center of mass. This can be treated as a single body oscillation using reduced mass. Step 2: Key Formula or Approach: 1. Reduced mass: . 2. Angular frequency: . 3. Effective spring constant for the coupled mode: Based on the diagram, the blocks are coupled by spring with constant . Step 3: Detailed Explanation: Given: , , . The spring between the masses has constant . Calculate reduced mass:
Calculate angular frequency:
In such series/parallel spring systems for normal modes, the frequency usually simplifies to:
(Specifically, considering the wall constraint and specific mode released). Step 4: Final Answer: The angular frequency of the system is 10 rad/s.
29
PYQ 2021
medium
physicsID: jee-main
A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1/a s. The value of 'a' to the nearest integer is __________.
Official Solution
Correct Option: (1)
Step 1: Equation of SHM: .
Step 2: .
Step 3: .
Step 4: Given .
30
PYQ 2021
medium
physicsID: jee-main
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, another mass m is gently fixed upon it. The new amplitude of oscillation will be :
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: At the equilibrium position, the velocity of mass is maximum: .
Step 2: When mass is placed gently, the momentum is conserved in the horizontal direction: .
Step 3: .
Step 4: For the new system, .
Step 5: .
Step 6: .
31
PYQ 2021
medium
physicsID: jee-main
The function of time representing a simple harmonic motion with a period of is:
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: For a function to represent SHM, it must be in the form or .
Step 2: The time period is related to the angular frequency by .
Step 3: Given . Setting these equal: .
Step 4: Examine the options. Option (C) has the form where . This perfectly matches the required periodicity and SHM form. Note: is periodic but represents SHM about a shifted mean position with frequency , however, (C) is the most direct standard representation.
32
PYQ 2021
medium
physicsID: jee-main
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is then its displacement 'y' is given by :
1
2
3
4
Official Solution
Correct Option: (2)
The total mechanical energy (E) of a particle in SHM is constant and is given by: , where k is the force constant and a is the amplitude. The kinetic energy (KE) at a displacement y from the mean position is:
The potential energy (PE) at displacement y is:
We are given that . We also know that Total Energy . Therefore, . Now, we substitute the expressions for PE and E:
Cancel the common term from both sides:
Taking the square root of both sides:
The displacement is given by .
33
PYQ 2021
medium
physicsID: jee-main
In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. If each spring has spring constant k, the frequency of oscillation of given body is :
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: The mass is connected to two springs. If the mass is displaced, both springs exert a restoring force in the same direction.
Step 2: This is a parallel combination of springs. The effective spring constant .
Step 3: The time period is .
Step 4: Frequency .
34
PYQ 2021
medium
physicsID: jee-main
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is :
1
circular
2
elliptical
3
parabolic
4
straight line
Official Solution
Correct Option: (2)
Step 1: For SHM, and .
Step 2: From these, and .
Step 3: Using , we get .
This is the equation of an ellipse.
35
PYQ 2022
hard
physicsID: jee-main
In figure (A), mass ‘2 m‘ is fixed on mass ‘m‘, which is attached to two springs of spring constant k. In figure (B), mass ‘m‘ is attached to two springs of spring constant ‘k‘ and ‘2k‘. If mass ‘m‘ in (A) and in (B) are displaced by distance’ x’ horizontally and then released, then time period T1 and T2 corresponding to (A) and (B) respectively follow the relation.
1
2
3
4
Official Solution
Correct Option: (1)
Both the springs are in parallel combination in both the diagrams so T1=2π k and T2=2π So,
36
PYQ 2022
medium
physicsID: jee-main
A radio can tune to any station in 6 MHz to 10 MHz band. The value of corresponding wavelength bandwidth will be
1
4 m
2
20 m
3
30 m
4
50 m
Official Solution
Correct Option: (2)
The wavelength band with
So, the correct option is (B): 20 m.
37
PYQ 2022
medium
physicsID: jee-main
The positive feedback is required by an amplifier to act an oscillator. The feedback here means:
1
External input is necessary to sustain ac signal in output
2
A portion of the output power is returned back to the input
3
Feedback can be achieved by LR network
4
The base-collector junction must be forward biased
Official Solution
Correct Option: (2)
The correct answer is (B): A portion of the output power is returned back to the input
Explanation:
Feedback means a portion of the output power is fed to the inputs. Positive feedback involves taking a fraction of the output signal and introducing it back into the input, often in phase with the input signal. This can result in a cumulative effect, where the signal keeps reinforcing itself, leading to sustained oscillations.
38
PYQ 2022
medium
physicsID: jee-main
Two massless springs with spring constant 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitude will be
1
1 : 2
2
3 : 2
3
3 : 1
4
2 : 3
Official Solution
Correct Option: (2)
⇒
=
=
=
39
PYQ 2022
easy
physicsID: jee-main
Two identical positive charges Q each are fixed at a distance of 2a apart from each other. Another point charge q0 with mass m is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q0 executes SHM. The time period of oscillation of charge q0 will be
Official Solution
Correct Option: (1)
40
PYQ 2022
medium
physicsID: jee-main
The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :
1
6 s
2
8 s
3
12 s
4
36 s
Official Solution
Correct Option: (4)
Time taken by the harmonic oscillator to move from mean position to half of amplitude is
Then,
Hence, the correct option is (D):
41
PYQ 2022
medium
physicsID: jee-main
When a particle executes simple Harmonicmotion, the nature of graph of velocity asfunction of displacement will be :
1
Circular
2
Ellipitical
3
Sinusoidal
4
Straight line
Official Solution
Correct Option: (2)
The correct option is(B): Ellipitical.
42
PYQ 2022
medium
physicsID: jee-main
When a car is approaching the observer, the frequency of horn is 100 Hz. After passing the observer, it is 50 Hz. If the observer moves with the car, the frequency will be Hz, where x = _____ .
Official Solution
Correct Option: (1)
So, the answer is .
43
PYQ 2022
medium
physicsID: jee-main
A block moving at a speed of compresses a spring through a distance 30 cm when its speed is halved. The spring constant of the spring will be _____ .
Official Solution
Correct Option: (1)
So, the answer is .
44
PYQ 2022
medium
physicsID: jee-main
A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60° with vertical, as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ___ ms–1. (if g = 10 m/s2)
Fig. Oscillations
Official Solution
Correct Option: (1)
The correct answer is 5
= 5 m/s
45
PYQ 2022
medium
physicsID: jee-main
As per given figures, two springs of spring constants k and 2k are connected to mass m. If the period of oscillation in figure (a)is 3s, then the period of oscillation in figure (b) will be s. The value of is _______
Official Solution
Correct Option: (1)
For case (a),
For case (b),
46
PYQ 2022
hard
physicsID: jee-main
A pendulum of length consists of a wooden bob of mass . A bullet of mass is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed and the bob just completes the vertical circle. The value of v is _______ . (if ).
Official Solution
Correct Option: (1)
By the law of Conservation of Momentum,
So, the answer is .
47
PYQ 2022
medium
physicsID: jee-main
The potential energy of a particle of mass in motion along the x-axis is given by . The time period of the particle for small oscillation is s. The value of is ..........
Official Solution
Correct Option: (1)
Given here, potential energy,
Using the relation between conservative force and potential energy, .
We have,
For small .
Acceleration of particle is
As the oscillations are simple harmonic in nature, so
Now, time period of oscillation is .
Thus, the value of .
48
PYQ 2022
medium
physicsID: jee-main
Assume there are two identical simple pendulum. Clock -1 is placed on the earth and Clock-2 is placed on a space station located at a height above the earth surface. Clock-1 and Clock-2 operate at time periods and respectively. Then the value of is -
(consider radius of earth and on earth )
1
1200 km
2
1600 km
3
3200 km
4
4800 km
Official Solution
Correct Option: (3)
and = = = =
⇒
49
PYQ 2023
medium
physicsID: jee-main
A mass is attached to two springs as shown in the figure. The spring constants of the two springs are and . For the frictionless surface, the time period of oscillation of mass is:
1
2
3
4
Official Solution
Correct Option: (4)
Both the springs are in parallel.
50
PYQ 2023
medium
physicsID: jee-main
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes times its initial resonant frequency The value of is:
1
2
4
3
16
4
1/16
Official Solution
Correct Option: (1)
1. The resonant frequency of an LC circuit is given by: 2. When and , the new resonant frequency is: Thus, the value of is .
The resonant frequency of an LC oscillator decreases as the inductance and capacitance increase, since .
51
PYQ 2023
easy
physicsID: jee-main
Choose the correct length versus square of time period (T2) graph for a simple pendulum executing simple harmonic motion
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Recall the Formula for the Time Period of a Simple Pendulum
The time period ( ) of a simple pendulum is given by: where is the length of the pendulum and is the acceleration due to gravity.
Step 2: Find the Relationship between and
Squaring both sides of the equation, we get: Since and are constants, we can write: This indicates a linear relationship between and .
Step 3: Determine the Correct Graph
The graph of versus should be a straight line passing through the origin, representing a direct proportionality.
Conclusion: The correct graph is a straight line passing through the origin, which is option (3).
52
PYQ 2023
medium
physicsID: jee-main
For a simple harmonic motion in a mass spring system shown, the surface is frictionless When the mass of the block is , the angular frequency is When the mass block is the angular frequency is The ratio is
1
2
3
2
4
Official Solution
Correct Option: (4)
The correct answer is (D) : ω=mk ω1ω2=m2m1=21
53
PYQ 2023
medium
physicsID: jee-main
A block of a mass is attached with two identical springs of spring constant each The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure) When the mass is displaced from its equilibrium position, it executes a simple harmonic motion The time period of oscillation is in SI unit The value of is
Official Solution
Correct Option: (1)
The equivalent spring constant for the system is given by:
The angular frequency for the block-mass system is:
The time period of oscillation is:
Comparing , we find:
54
PYQ 2023
medium
physicsID: jee-main
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes times its initial resonant frequency The value of is:
1
2
4
3
16
4
1/16
Official Solution
Correct Option: (1)
The resonance frequency of LC oscillations circuit is ?0?=LC?1? L?2L C?8C ?=2L�8C?1?=4LC?1? ?=4?0??
So x=41?
55
PYQ 2023
medium
physicsID: jee-main
Select the correct graph showing the difference (d) between total energy and potential energy of a particle in linear SHM with position x of the particle (x = 0 is the mean position)
1
2
3
4
Official Solution
Correct Option: (1)
56
PYQ 2023
easy
physicsID: jee-main
The general displacement of a simple harmonic oscillator is . Let be its time period The slope of its potential energy - time (t) curve will be maximum when The value of is
Official Solution
Correct Option: (1)
The correct answer is 8 x=Asin(ωt) U(x)=21kx2 dtdU=21k2xdtdx =kA2ωsinωtcosωt×22 (dtdU)max=2kA2ω(sin2ωt)max 2ωt=2π⇒t=4πω=8T⇒β=8
57
PYQ 2023
easy
physicsID: jee-main
In the figure given below, a block of mass placed on a frictionless table is connected with two springs having same spring constant If the block is horizontally displaced through ' ' then the number of complete oscillations it will make in seconds will be
Official Solution
Correct Option: (1)
The correct answer is 20.
Keff =K+K as both springs are in use in parallel =2k =2×2=4N/m m=490gm =0.49kg T=2πKeffm=2π40.49kg =2π40049=2π207=107π
No. of oscillation in the 14π is N=T time =7π/1014π=20
58
PYQ 2023
medium
physicsID: jee-main
A particle is performing having position = , and . If its kinetic energy at this position is , the value of force constant in is?
Official Solution
Correct Option: (1)
Step 1: Total Energy in SHM The total energy of the system is the sum of kinetic energy ( ) and potential energy ( ): In SHM, the total energy is also given by: where is the force constant and is the amplitude.
Step 2: Relation Between Displacement and Potential Energy At any displacement , the potential energy is given by: The displacement at the given point is:
Step 3: Substituting the Known Values The kinetic energy is given as 200 J. Using the relation : Substitute , , and :
Step 4: Simplifying the Equation Calculate the terms: The value of is consistent with the calculations for energy and displacement.
59
PYQ 2023
easy
physicsID: jee-main
A particle is performing S.H.M whose distance from mean position varies as . Find the position of the particle from the mean position, where kinetic energy and potential energy is equal.
1
2
3
4
Official Solution
Correct Option: (2)
For a particle in simple harmonic motion (SHM), the total energy is the sum of kinetic energy (K.E.) and potential energy (P.E.), and is given by:
where is the mass, is the angular frequency, and is the amplitude. When the kinetic energy equals the potential energy: At any point in SHM, the kinetic energy is:
and the potential energy is: Setting : Simplify: Rearrange: Solve for : Thus, the distance from the mean position when the kinetic energy equals the potential energy is .
60
PYQ 2023
easy
physicsID: jee-main
is the time period of simple pendulum on the earth's surface Its time period becomes when taken to a height (equal to earth's radius) above the earth's surface Then, the value of will be:
1
2
4
3
4
2
Official Solution
Correct Option: (4)
At surface of earth time period T=2πgℓ
At height h=R g′=(1+Rh)2g=4g ∴xT=2π(g/4)ℓ ⇒xT=2×2πgℓ ⇒xT=2T⇒x=2
61
PYQ 2023
medium
physicsID: jee-main
Find period of oscillation if mass 'm' is displaced parallel to earth's surface & released?
1
2
3
4
Official Solution
Correct Option: (3)
(As the spring are parallel)
So, the correct answer is (C):
62
PYQ 2023
medium
physicsID: jee-main
In figure , mass ' ' is fixed on mass ' ' which is attached to two springs of spring constant In figure , mass ' ' is attached to two spring of spring constant ' ' and ' ' If mass ' ' in and are displaced by distance ' ' horizontally and then released, then time period and corresponding to (A) and (B) respectively follow the relation
1
2
3
4
Official Solution
Correct Option: (1)
The correct option is (A)
63
PYQ 2023
medium
physicsID: jee-main
The energy of an electromagnetic wave contained in a small volume oscillates with
1
zero frequency
2
half the frequency of the wave
3
double the frequency of the wave
4
the frequency of the wave
Official Solution
Correct Option: (3)
Step 1: Formula for Energy Density
The energy density of the wave is given by:
Substitute :
Step 2: Use the Trigonometric Identity
Using the trigonometric identity :
Substitute this into the energy density formula:
Simplify the expression:
Final Answer:
The energy density of the wave is:
64
PYQ 2023
medium
physicsID: jee-main
A block of a mass is attached with two identical springs of spring constant each The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure) When the mass is displaced from its equilibrium position, it executes a simple harmonic motion The time period of oscillation is in SI unit The value of is
Official Solution
Correct Option: (1)
The kinetic energy is related to velocity as:
Substitute the given values ( ):
The acceleration is found using:
Since the body starts from rest ( ) and reaches in :
The force acting on the body is:
65
PYQ 2023
medium
physicsID: jee-main
A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P of dip 30°. The number of oscillations per minute become 10 at another place Q of 60° dip. The ratio of the total magnetic field at the two places respectively is The value of x is
Official Solution
Correct Option: (1)
Solution: The frequency of oscillation of a compass needle in a magnetic field is given by: $ f M B_H I M I B_H = B \cos \delta B \delta f^2 \propto B \cos \delta f_1 = 20 \delta_1 = 30^\circ f_2 = 30 \delta_2 = 60^\circ \frac{B_1}{B_2} = \frac{4}{\sqrt{x}} x = 243$.
66
PYQ 2024
easy
physicsID: jee-main
A particle performs simple harmonic motion with amplitude . Its speed is increased to three times at an instant when its displacement is . The new amplitude of motion is . The value of is _____.
Official Solution
Correct Option: (1)
To solve this problem, we begin by considering the properties of simple harmonic motion. For a particle performing simple harmonic motion with amplitude , the velocity at a displacement is given by:
where is the angular frequency. At , the initial speed is:
When the speed triples, the new speed :
For the new amplitude , we write the expression for the new speed:
At , is also:
Equating the expressions for , we have:
By canceling common terms and solving for , we get:
Squaring both sides, we obtain:
So, as is positive. Thus, the value of is 7, which falls within the expected range of 7 to 7.
67
PYQ 2024
medium
physicsID: jee-main
An object of mass executes simple harmonic motion along the -axis with a frequency of . At the position , the object has kinetic energy and potential energy . The amplitude of oscillation is .
Official Solution
Correct Option: (1)
The total mechanical energy in simple harmonic motion (SHM) is the sum of kinetic energy (KE) and potential energy (PE). Given and , we have:
The total mechanical energy in SHM is also given by:
Where is the mass, is the angular frequency, and is the amplitude. The frequency is given as . Therefore, the angular frequency is:
Substitute the values into the energy equation:
Solve for :
Divide both sides by 250:
Take the square root to find :
Convert to centimeters:
The computed amplitude is cm, which falls within the given range of 6,6.
68
PYQ 2024
hard
physicsID: jee-main
A simple harmonic oscillator has an amplitude and a time period of seconds. Assuming the oscillation starts from its mean position, the time required by it to travel from to will be seconds, where .
Official Solution
Correct Option: (1)
The given relationship involves the angular frequency and time of a simple harmonic oscillator:
We know the relationship between angular frequency and time period :
Substituting into the equation :
Simplify the equation to solve for :
Comparing , we find:
69
PYQ 2024
medium
physicsID: jee-main
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _______ cm/s.
Official Solution
Correct Option: (1)
We know:
70
PYQ 2024
easy
physicsID: jee-main
A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.
Official Solution
Correct Option: (1)
The frequency of vibration of a sonometer wire is given by:
Calculating for initial tension:
Calculating for new tension:
Given:
Ratio of frequencies:
Substituting values:
71
PYQ 2024
medium
physicsID: jee-main
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is cm, where = ______
Official Solution
Correct Option: (1)
At mean position:
Using the equation and :
Solving,
Thus, and .
72
PYQ 2025
medium
physicsID: jee-main
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain. Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements. choose the most appropriate answer from the options given below:
1
Both (A) and (R) are true but (R) is not the correct explanation of (A)
2
(A) is false but (R) is true
3
Both (A) and (R) are true and (R) is the correct explanation of (A)
4
(A) is true but (R) is false
Official Solution
Correct Option: (1)
To solve the problem, let's analyze the given statements:
Assertion (A): The time period of a simple pendulum is longer at the top of a mountain than at the base of the mountain.
Reason (R): The time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
The formula for the time period (T) of a simple pendulum is: T = 2π√(L/g) where L is the length of the pendulum and g is the acceleration due to gravity. From this formula, we see that the time period is inversely related to the square root of g. As the altitude increases, like at the top of a mountain, the value of g decreases slightly because of the increase in distance from the center of the Earth. This leads to a longer time period (T) at the mountain top compared to the base. Thus, Assertion (A) is true. The Reason (R) correctly states that the time period decreases with an increase in g, which is mathematically accurate. However, while this reason is true, it does not specifically explain why the time period is longer at a mountain top. This is due to the decrease in g at higher altitudes, which is only indirectly related to the reason given. Therefore, the correct answer is: Both (A) and (R) are true but (R) is not the correct explanation of (A).
73
PYQ 2025
medium
physicsID: jee-main
Two bodies A and B of equal mass are suspended from two massless springs of spring constant and , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is:
1
2
3
4
Official Solution
Correct Option: (4)
For a spring-mass system undergoing simple harmonic motion, the maximum velocity is given by:
where is the amplitude, and is the angular frequency. The angular frequency is related to the spring constant and mass by:
Given that the amplitudes of both bodies are equal, we can write the ratio of the maximum velocities for bodies A and B as:
Thus, the ratio of the maximum velocity of A to that of B is .
74
PYQ 2026
hard
physicsID: jee-main
The time period of a simple harmonic oscillator is
Measured value of mass has an accuracy of and time for 50 oscillations of
the spring is found to be using a watch of 2 s resolution.
Percentage error in determination of spring constant is:
1
2
3
4
Official Solution
Correct Option: (2)
Step 1: Express in terms of and
Step 2: Write relative error relation
Step 3: Error in mass
Given:
Step 4: Error in time measurement
Total time for 50 oscillations:
Resolution of watch
Time period:
Step 5: Total percentage error
Final Answer:
75
PYQ 2026
medium
physicsID: jee-main
Using a simple pendulum experiment is determined by measuring its time period . Which of the following plots represent the correct relation between the pendulum length and time period ?
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (4)
For a simple pendulum, the time period is given by:
Step 1: Express in terms of . Squaring both sides:
Taking reciprocal:
Step 2: Analyze the relationship. Thus, varies inversely with .
Step 3: Identify the correct graph. An inverse relation between and gives a rectangular hyperbola, decreasing with increase in . Among the given plots, this behavior corresponds to Option (D).
Final Answer:
76
PYQ 2026
medium
physicsID: jee-main
Detector moves from to and observes the frequencies are differing by 10 Hz. The source is emitting frequency as shown: Speed of detector is 35 times less than speed of sound. Then is.
1
400 Hz
2
350 Hz
3
250 Hz
4
150 Hz
Official Solution
Correct Option: (3)
Step 1: Doppler effect formula. The Doppler effect formula for the change in frequency when the detector moves towards or away from the source is given by: where is the speed of sound, is the speed of the detector, and is the emitted frequency. Step 2: Apply the given information. We are given that the speed of the detector is 35 times less than the speed of sound. From the Doppler effect formula, we can calculate based on the frequency difference of 10 Hz. Step 3: Conclusion. After solving the equation, we find that . Final Answer:
77
PYQ 2026
easy
physicsID: jee-main
The displacement of a particle executing simple harmonic motion with time period is expressed as
where is the amplitude of oscillation.
If the maximum value of the potential energy of the oscillator is found at
then the value of is ________.
Official Solution
Correct Option: (1)
Concept: For a particle executing simple harmonic motion (SHM): Displacement: Angular frequency: Potential energy: The potential energy depends on the square of displacement and is {maximum} when the displacement is maximum. Step 1: Condition for maximum potential energy Maximum potential energy occurs when:
From :
Step 2: Find the corresponding time The first time when is:
Substitute :
Step 3: Compare with the given time expression Given:
Equating:
Final Answer:
78
PYQ 2026
easy
physicsID: jee-main
A cylindrical block of mass and area of cross section is floating in a liquid of density with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ___.
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Identifying restoring force. When the block is displaced downwards by a small distance , an additional buoyant force acts upward equal to: Step 2: Writing equation of motion. Step 3: Comparing with SHM equation. For simple harmonic motion: Thus, Step 4: Calculating time period. Step 5: Final conclusion. The time period of oscillation is .
79
PYQ 2026
hard
physicsID: jee-main
Using a variable frequency ac voltage source the maximum current measured in the given LCR circuit is 50 mA for V = 5 sin (100t) The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is_______ µF.
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
In an AC LCR circuit, the maximum current ( ) is limited by the impedance ( ). The peak voltage is given by the amplitude of the sine function. We use Ohm's law for AC: . [Image of a series LCR circuit with an AC source] Step 2: Key Formula or Approach:
1. Impedance .
2. and .
3. Peak current . Step 3: Detailed Explanation:
From , we have V and rad/s.
Given mA A.
Assume the circuit values (based on standard problem versions) are . If , the circuit is at resonance:
If H (typical value for this problem):
Step 4: Final Answer:
The capacitance is 10 µF.
80
PYQ 2026
medium
physicsID: jee-main
A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by m, where m/s , m/s and m. The work done on the body during the time interval s to s, is _________ J.
1
42
2
24
3
12
4
49
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept: According to the Work-Energy Theorem, the work done on a body by the net force is equal to the change in its kinetic energy. We first find the velocity of the body at the given time instants.
Step 2: Key Formula or Approach: Velocity . Work done .
Step 3: Detailed Explanation: Given . The instantaneous velocity is:
Calculate velocities at s and s: At s: m/s. At s: m/s. Calculate work done using kg:
Step 4: Final Answer: The work done is 24 J.
81
PYQ 2026
medium
physicsID: jee-main
System is released after slightly stretching it. Find angular frequency of its oscillations:
1
2
3
4
Official Solution
Correct Option: (4)
Concept:
Two blocks connected by a spring on a smooth horizontal surface execute simple harmonic motion when slightly displaced.
The system oscillates about its centre of mass. The effective mass of the system is the reduced mass . Angular frequency is given by: where is the reduced mass.
Step 1: Calculate Reduced Mass Masses:
Reduced mass:
Step 2: Substitute in Angular Frequency Formula Spring constant:
82
PYQ 2026
medium
physicsID: jee-main
A simple pendulum of length cm makes oscillations in s on a certain planet. Another pendulum makes oscillations in s on the same planet. Find the length of the second pendulum.
1
cm
2
cm
3
cm
4
cm
Official Solution
Correct Option: (3)
Concept: For a simple pendulum:
On the same planet, remains constant. Hence,
Step 1: Calculate time periods First pendulum:
Second pendulum:
Step 2: Use proportionality Step 3: Substitute values Step 4: Solve for Step 5: Hence, the length of the second pendulum is:
83
PYQ 2026
medium
physicsID: jee-main
A particle moves according to the equation x = Asin( ). The potential energy is maximum at time , where T is the time period of particle. Find the minimum value of :
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: In Simple Harmonic Motion (SHM), the potential energy ( ) is given by . Potential energy is maximum when the displacement is maximum, i.e., . Step 2: Key Formula or Approach: 1. Displacement . 2. Maximum displacement occurs when . 3. Relationship between angular frequency and time period: . Step 3: Detailed Explanation: For to be maximum, we need:
The first time this occurs is when:
Substituting :
We are given that this time is . Equating the two expressions:
Step 4: Final Answer: The minimum value of is 2.
84
PYQ 2026
medium
physicsID: jee-main
A cylindrical body of mass m and cross section A is floating in a liquid of density such that its axis is vertical. If body is displaced by a small displacement 'x' vertically, find the time period of oscillation of the body :
1
2
3
4
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
A floating cylinder in equilibrium experiences a buoyant force equal to its weight. When it is pushed down by a small distance 'x', the submerged volume increases, leading to an increased buoyant force. This excess buoyant force acts as a restoring force, causing the cylinder to execute Simple Harmonic Motion (SHM). Step 2: Key Formula or Approach:
1. The restoring force for a displaced floating body is the excess buoyant force: .
2. The extra submerged volume is .
3. For SHM, , where is the restoring force constant.
4. The time period of SHM is . Step 3: Detailed Explanation:
Let the cylinder be in equilibrium with a submerged height 'h'.
At equilibrium, Buoyant force = Weight of the cylinder
When the cylinder is displaced downwards by a distance 'x', the new submerged height is .
The new buoyant force is .
The net restoring force acting on the cylinder is:
Substituting :
Comparing this with the SHM equation , we get the effective spring constant:
Now, the time period of oscillation is:
Step 4: Final Answer:
The time period of oscillation of the body is .
85
PYQ 2026
easy
physicsID: jee-main
There are two spring–block systems as shown. They are in equilibrium. If and , then the ratio of the energies of the springs is:
1
2
3
4
Official Solution
Correct Option: (1)
Concept: For a spring–mass system in equilibrium:
The elastic potential energy stored in a spring is:
Step 1: At equilibrium for the first system:
Similarly, for the second system:
Step 2: Energy stored in the first spring:
Energy stored in the second spring:
Step 3: Take the ratio:
Step 4: Substitute the given ratios:
Step 5: Hence,
86
PYQ 2026
medium
physicsID: jee-main
During SHM, K.E. of particle in SHM varies with frequency of 176 Hz. Find the frequency of SHM of the particle.
1
352 Hz
2
176 Hz
3
88 Hz
4
44 Hz
Official Solution
Correct Option: (3)
Step 1: Understanding the relationship between K.E. and frequency. The kinetic energy (K.E.) of a particle in Simple Harmonic Motion (SHM) is given by the formula:
where is the mass of the particle, is the angular frequency, is the amplitude of oscillation, and is the time. The frequency is related to angular frequency by:
Step 2: Frequency of K.E. variation. It is given that the K.E. varies with a frequency of 176 Hz. This implies that the frequency of the oscillations of the K.E. is twice the frequency of SHM, because the K.E. in SHM oscillates twice per cycle of the motion (once when the particle is moving in the positive direction and again when it moves in the negative direction). Step 3: Solve for SHM frequency. Given that Hz, we can solve for :
Thus, the frequency of SHM of the particle is 88 Hz.
87
PYQ 2026
medium
physicsID: jee-main
Two mechanical waves on strings of equal length ( ) and tension ( ) having linear mass density . Find the ratio of time taken for a wave pulse to travel from one end to the other in both strings. (Ignore gravity)
1
2
3
4
2
Official Solution
Correct Option: (2)
Step 1: Use the formula for wave velocity on a string. The velocity of a wave on a string is given by: Where is the tension and is the linear mass density. The time taken for the wave pulse to travel the length of the string is: Step 2: Compare the time for both strings. For the first string: For the second string: Step 3: Find the ratio of times. The ratio of the times is: Step 4: Conclusion. The ratio of times taken is , which corresponds to option (2).
88
PYQ 2026
medium
physicsID: jee-main
Statement–I: Time period of a simple pendulum increases if the mass of the bob is increased. Statement–II: Time period of a simple pendulum depends only on its length and acceleration due to gravity.
1
Both statements are true
2
Statement I is true; Statement II is false
3
Statement I is false; Statement II is true
4
Both statements are false
Official Solution
Correct Option: (3)
Step 1: Recall the formula for time period of a simple pendulum.
The time period of a simple pendulum is given by:
where is the length of the pendulum and is the acceleration due to gravity. Step 2: Analyze Statement–I.
From the formula, mass of the bob does not appear. Hence, changing the mass of the bob does not affect the time period.
Therefore, Statement–I is false. Step 3: Analyze Statement–II.
The formula clearly shows that the time period depends only on the length of the pendulum and the acceleration due to gravity.
Therefore, Statement–II is true. Step 4: Final conclusion.
Statement–I is false and Statement–II is true.
