AP-EAPCET SERIES

Physics

Simple Harmonic Motion

31 previous year questions.

Volume: 31 Ques

Yield: High

High-Yield Trend

2025

2024

2023

Chapter Questions
31 MCQs

PYQ 2023

medium

physics ID: ap-eapce

An object of mass 3 kg is executing simple harmonic motion with an amplitude of . If the kinetic energy of the object when it crosses the mean position is 6 J, the time period of oscillation of the object is:

1 s

2 s

3 s

4 s

PYQ 2023

medium

physics ID: ap-eapce

Two simple pendulums of lengths 1 m and 2 m are in phase at the mean position at a certain instant of time. The pendulums will be again in phase at a time of, where is the time period of the shorter pendulum. Then ?

4 m

6 m

9 m

12 m

PYQ 2023

medium

physics ID: ap-eapce

The equation of motion of a particle executing simple harmonic motion is given by: where is in meters and is in seconds. The time period of oscillation of the particle is:

PYQ 2023

medium

physics ID: ap-eapce

A block A of mass 1.6 kg is executing SHM with amplitude of 10 cm. A block B of mass 0.9 kg is kept on A. When A passes through its equilibrium, the amplitude of oscillation of the system of block A and B is

4 cm

8 cm

12 cm

16 cm

PYQ 2023

medium

physics ID: ap-eapce

The equation of a particle executing simple harmonic motion is . The maximum velocity of the particle is:

PYQ 2023

medium

physics ID: ap-eapce

The equation of motion of a particle executing simple harmonic motion is given by, where x is in metres and t is in seconds. The ratio of the potential and kinetic energies of the particle at time is

PYQ 2023

medium

physics ID: ap-eapce

Two particles execute simple harmonic motion along the same straight line with same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distan of times the amplitude from their common mean position. The phase differen between the two particles is

1

2

3

4

Official Solution

Correct Option: $(4)$

$Let both particles have amplitude . Let, and . At the instant of crossing, . Then,; thus, . Also, they are moving in opposite directions velocities are opposite: . So, . Thus, the phase difference is .$

08

PYQ 2023

medium

physics ID: ap-eapce

$The acceleration of a particle executing simple harmonic motion at a distance of 10 cm from the mean position is 0.4 ms . If the amplitude of the particle is 30 cm, the maximum acceleration of the particle is$

1

$0.6 ms$

2

$0.9 ms$

3

$1.2 ms$

4

$1.5 ms$

Official Solution

Correct Option: $(3)$

$In simple harmonic motion (SHM), the acceleration at a displacement from the mean position is given by: where is the angular frequency. Given: -, - . Substituting the values: The maximum acceleration in SHM occurs at the amplitude, and is given by: Given the amplitude, substitute : So, the maximum acceleration of the particle is 1.2 ms .$

09

PYQ 2023

medium

physics ID: ap-eapce

$A body thrown vertically upwards with a velocity of 20 ms from the surface of a planet reaches back the surface of the planet in a time of 2 s. The time period of a simple pendulum of length m on this planet is$

1

$4 s$

2

$3 s$

3

$2 s$

4

$1 s$

Official Solution

Correct Option: $(3)$

$First, determine the acceleration due to gravity on the planet. The body is thrown upwards with an initial velocity and returns to the surface in 2 s. The time to reach the maximum height (where velocity becomes zero) is half of the total time, so . Using the equation of motion: Now, calculate the time period of the simple pendulum. The time period of a simple pendulum is given by: Given the length of the pendulum, and, substitute the values: So, the time period of the simple pendulum is 2 s .$

10

PYQ 2023

medium

physics ID: ap-eapce

$The equation of motion of a particle executing simple harmonic motion is given by, where is in meters and is in seconds. The maximum acceleration of the particle in m/s² is$

1

$9$

2

$12$

3

$18$

4

$6$

Official Solution

Correct Option: $(3)$

$The maximum acceleration in SHM is given by, where is the angular frequency and is the amplitude. From the given equation, the angular frequency and the amplitude . Thus, the maximum acceleration is: Therefore, the maximum acceleration is 18 m/s².$

11

PYQ 2023

medium

physics ID: ap-eapce

$Under the action of a force where F is in Newton and y is in meters, an object of mass 3 kg executes simple harmonic motion. If the velocity of the object at the mean position is 2.5 ms, the maximum acceleration of the object is$

1

$ms$

2

$ms$

3

$ms$

4

$ms$

Official Solution

Correct Option: $(4)$

$The force acting on the object is . According to Newton's second law,, where is the mass and is the acceleration. So, . The acceleration of the object is . Given mass kg, the acceleration is . For simple harmonic motion, the acceleration is given by, where is the angular frequency. Comparing the two expressions for acceleration, we have, so rad/s. The velocity of the object in SHM is given by, where is the amplitude of the motion. At the mean position,, and the velocity is maximum, . Given that the velocity at the mean position is ms . So, . The amplitude m. The maximum acceleration in SHM occurs at the extreme positions () and is given by . Substituting the values of and : ms .$

12

PYQ 2023

medium

physics ID: ap-eapce

$A particle is executing simple harmonic motion. If the minimum time taken by the particle to move from extreme position to half of the amplitude is, and the minimum time taken by the particle to move from mean position to half of the amplitude is, then$

1

2

3

4

Official Solution

Correct Option: $(3)$

$In simple harmonic motion, the time taken to move from extreme position to half the amplitude is, which corresponds to a quarter of the period of the oscillation. The time taken to move from mean position to half of the amplitude is, which corresponds to half the period . Therefore, we can conclude: Thus, the correct answer is:$

13

PYQ 2023

medium

physics ID: ap-eapce

$A particle initially at the mean position is executing simple harmonic motion with an angular frequency . The ratio of the distances travelled by the particle in the first second and second second is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$In SHM, distance travelled in time from mean is related to the sine function. Use the formula for displacement: At, At, Hence distance in 1st second: Distance in 2nd second: Ratio =$

14

PYQ 2023

medium

physics ID: ap-eapce

$When a body of mass 8 kg is attached to a spring balance, the reading of the balance is 20 cm. Instead of 8 kg, if another body of mass M is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is s, then the value of M is (Acceleration due to gravity = 10 ms)$

1

$kg$

2

$kg$

3

$kg$

4

$kg$

Official Solution

Correct Option: $(1)$

$When a mass is attached to a spring balance, the spring extends until the spring force balances the gravitational force:, where is the spring constant and is the extension. In the first case, mass kg and extension cm m. N/m When a mass is suspended and oscillates vertically, the time period of oscillation for a mass-spring system is given by . Given time period s. Divide both sides by : Divide both sides by 2: Square both sides: Solve for : kg The value of M is 4 kg.$

15

PYQ 2024

medium

physics ID: ap-eapce

$A body is executing simple harmonic motion. At a displacement, its potential energy is, and at a displacement, its potential energy is . The potential energy at a displacement is:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Understanding Potential Energy in SHM The potential energy in SHM is given by: where is the force constant. Step 2: Applying to Given Displacements For two different displacements: For total displacement : Step 3: Using the Energy Sum Property Since energy follows a quadratic relationship: Thus, the correct answer is option (2).$

16

PYQ 2025

medium

physics ID: ap-eapce

$If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively and, then the time period of oscillation of the particle is$

1

2

3

$% Image seems to cut off option 3, assuming it was 2s$

4

Official Solution

Correct Option: $(1)$

$For a particle executing Simple Harmonic Motion (SHM), let A be the amplitude and be the angular frequency. Maximum velocity . Maximum acceleration . Given: Divide equation (2) by equation (1): The time period of oscillation is related to by . This matches option (1).$

17

PYQ 2025

medium

physics ID: ap-eapce

$A particle is executing simple harmonic motion with amplitude . At a distance from the mean position, when the particle is moving towards the extreme position, it receives a blow in the direction of motion that instantaneously doubles its velocity. The new amplitude of the particle is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Energy Conservation in SHM The total mechanical energy in simple harmonic motion is: At displacement, the energy distribution is: Since the velocity is doubled instantly, kinetic energy increases by a factor of 4: Equating energies: Solving for : Conclusion Thus, the correct answer is:$

18

PYQ 2025

medium

physics ID: ap-eapce

$A particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is 1.5 s, then the minimum time at which the ratio of the kinetic and total energies of the particle becomes 3:4 is$

1

$1/4 s$

2

$1/12 s$

3

$1/8 s$

4

$1/6 s$

Official Solution

Correct Option: $(3)$

$Let the equation of SHM be, where is the amplitude and is the angular frequency. Since the particle starts from the mean position, the initial phase is 0. Kinetic energy . Total energy . The ratio of kinetic to total energy is . We are given that, so . . The time period s, so ., so s. This is the minimum time.$

19

PYQ 2025

medium

physics ID: ap-eapce

$A body of mass is attached to the lower end of a vertically suspended spring of force constant . If another body of mass moving vertically upward hits the suspended body with a velocity and is embedded in it, then the frequency of the oscillation is}$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Total mass after collision Frequency$

20

PYQ 2025

medium

physics ID: ap-eapce

$A particle is executing simple harmonic motion with amplitude . At a distance from the mean position, when the particle is moving towards the extreme position, it receives a blow in the direction of motion that instantaneously doubles its velocity. The new amplitude of the particle is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Energy Conservation in SHM The total mechanical energy in simple harmonic motion is: At displacement, the energy distribution is: Since the velocity is doubled instantly, kinetic energy increases by a factor of 4: Equating energies: Solving for : Conclusion Thus, the correct answer is:$

21

PYQ 2025

medium

physics ID: ap-eapce

$If the displacement (in cm) of a particle executing simple harmonic motion is given by the equation: then the amplitude of the particle is:$

1

2

3

4

Official Solution

Correct Option: $(4)$

$Step 1: Rewrite in standard form The given equation is: Step 2: Use trigonometric identity We can express this as: where amplitude Step 3: Calculate amplitude Here, :$

22

PYQ 2025

medium

physics ID: ap-eapce

$The kinetic energy of a particle executing simple harmonic motion at a displacement of from the mean position is . If the amplitude of the particle is, then the maximum force acting on the particle is}$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Kinetic energy in SHM is given by: Given,, Substitute: Maximum force$

23

PYQ 2025

medium

physics ID: ap-eapce

$The amplitude of a damped harmonic oscillator becomes 50% of its initial value in a time of 12 s. If the amplitude of the oscillator at a time of 36 s is of its initial amplitude, then the value of is$

1

2

3

4

Official Solution

Correct Option: $(2)$

$Step 1: Damped amplitude decreases exponentially: Step 2: Given: at, amplitude becomes : Step 3: At, Step 4: Percentage of initial amplitude: % Final Answer$

24

PYQ 2025

medium

physics ID: ap-eapce

$If the potential energy of a particle of mass moving along the x-axis is, then the speed of the particle is maximum at a position of:$

1

2

3

4

Official Solution

Correct Option: $(2)$

$The potential energy is given as: The force is the negative gradient of the potential energy: Calculating the derivative: The speed of the particle is maximum where the force is zero, which occurs at .$

25

PYQ 2025

medium

physics ID: ap-eapce

$A body of mass 1 kg is suspended from a spring of force constant . Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of and embedded in it. The amplitude of motion is$

1

$5 cm$

2

$15 cm$

3

$10 cm$

4

$8 cm$

Official Solution

Correct Option: $(1)$

$Let kg (suspended mass), . Let kg (hitting mass), (its initial velocity). Initially, mass M is suspended, so the spring is stretched by such that . m (assuming). This is the initial equilibrium position. Collision: The mass m hits M and embeds. This is a perfectly inelastic collision. Let be the velocity of the combined mass just after collision. By conservation of linear momentum (assuming collision time is very short, spring force impulse is negligible): (Initial velocity of M is 0 as it's suspended at rest). upwards. New system: Combined mass kg. Spring constant . The new equilibrium position for mass is where . m. This new equilibrium position is measured from the unstretched length of the spring. At the moment of collision, the system (mass M) was at its equilibrium position m. The collision occurs at this position. So, just after collision, the combined mass is at m from the unstretched position, and has velocity upwards. The SHM occurs about the new equilibrium position m. The displacement of the combined mass from its new equilibrium position at (just after collision) is: m. The negative sign means it is m above the new equilibrium position. The velocity at this displacement is . For SHM, the total energy is constant:, where A is the amplitude. Also,, where is displacement from new equilibrium. Angular frequency of new SHM: . The velocity and displacement from equilibrium are related to amplitude by . . . . More precisely: . This is very close to 5 cm. Let's verify the use of . If : . . m. . . . m cm. The result is consistently around 5 cm. So option (1) is correct.$

26

PYQ 2025

medium

physics ID: ap-eapce

$A body of mass kg attached to a spring of force constant N/m executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is:$

1

$s$

2

$s$

3

$s$

4

$s$

Official Solution

Correct Option: $(2)$

$Step 1: Formula for Time Period in SHM The time period of a simple harmonic oscillator is given by: where: - kg (mass of the body), - N/m (force constant of the spring). Step 2: Substituting Values Conclusion Thus, the correct answer is:$

27

PYQ 2025

medium

physics ID: ap-eapce

$On a smooth inclined plane, a block of mass is fixed to two rigid supports using two springs, each having spring constant, as shown in the figure. If the masses of the springs are neglected, then the period of oscillation of the block is$

1

2

3

4

Official Solution

Correct Option: $(1)$

$Each spring has spring constant, and they are connected in parallel to the block. Effective spring constant is: The formula for period of oscillation for mass-spring system is: Thus, the period of oscillation of the block is .$

28

PYQ 2025

medium

physics ID: ap-eapce

$A body of mass kg attached to a spring of force constant N/m executes simple harmonic motion on a frictionless horizontal surface. The time period of oscillation is:$

1

$s$

2

$s$

3

$s$

4

$s$

Official Solution

Correct Option: $(2)$

$Step 1: Formula for Time Period in SHM The time period of a simple harmonic oscillator is given by: where: - kg (mass of the body), - N/m (force constant of the spring). Step 2: Substituting Values Conclusion Thus, the correct answer is:$

29

PYQ 2025

medium

physics ID: ap-eapce

$The angular frequency of a block of mass 0.1 kg oscillating with the help of a spring of force constant 2.5 N m is:$

1

$0.2 rad s$

2

$5 rad s$

3

$10 rad s$

4

$2 rad s$

Official Solution

Correct Option: $(2)$

$Step 1: Recall the formula for angular frequency For a spring-mass system: Step 2: Substitute given values Given and :$

30

PYQ 2025

medium

physics ID: ap-eapce

$A body of mass 1 kg is suspended from a spring of force constant . Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of and embedded in it. The amplitude of motion is$

1

$5 cm$

2

$15 cm$

3

$10 cm$

4

$8 cm$

Official Solution

Correct Option: $(1)$

$Let kg (suspended mass), . Let kg (hitting mass), (its initial velocity). Initially, mass M is suspended, so the spring is stretched by such that . m (assuming). This is the initial equilibrium position. Collision: The mass m hits M and embeds. This is a perfectly inelastic collision. Let be the velocity of the combined mass just after collision. By conservation of linear momentum (assuming collision time is very short, spring force impulse is negligible): (Initial velocity of M is 0 as it's suspended at rest). upwards. New system: Combined mass kg. Spring constant . The new equilibrium position for mass is where . m. This new equilibrium position is measured from the unstretched length of the spring. At the moment of collision, the system (mass M) was at its equilibrium position m. The collision occurs at this position. So, just after collision, the combined mass is at m from the unstretched position, and has velocity upwards. The SHM occurs about the new equilibrium position m. The displacement of the combined mass from its new equilibrium position at (just after collision) is: m. The negative sign means it is m above the new equilibrium position. The velocity at this displacement is . For SHM, the total energy is constant:, where A is the amplitude. Also,, where is displacement from new equilibrium. Angular frequency of new SHM: . The velocity and displacement from equilibrium are related to amplitude by . . . . More precisely: . This is very close to 5 cm. Let's verify the use of . If : . . m. . . . m cm. The result is consistently around 5 cm. So option (1) is correct.$

31

PYQ 2025

hard

physics ID: ap-eapce

$A particle executes SHM with amplitude and time period . The time taken to move from equilibrium to is:$

1

2

3

4

Official Solution

Correct Option: $(3)$

$In simple harmonic motion (SHM), the displacement at time is given by: where is the angular frequency. Given that the particle moves from equilibrium (i.e., at) to in time : Therefore, Recall, so: However, this is the time taken to reach for the first time after starting at 0 (equilibrium). Note: The particle moves from 0 to within this time. Hence, the time taken is: This matches option (A). Wait, the options given in the question are different from the final result. The correct time from equilibrium to is . If your options differ, the answer is option (A) .$

About Simple Harmonic Motion - AP-EAPCET

Simple Harmonic Motion is a vital chapter for AP-EAPCET aspirants. Mastering the concepts covered in this chapter is essential for securing a top rank.

By rigorously practicing the previous year questions associated with this chapter, you can identify high-yield topics, understand the examiner's perspective, and boost your confidence during the actual exam.

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Simple Harmonic Motion

High-Yield Trend

Chapter Questions 31 MCQs

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About Simple Harmonic Motion - AP-EAPCET

Frequently Asked Questions

Why focus on Simple Harmonic Motion PYQs?

How to best use this analysis?

Chapter Questions
31 MCQs