Simple Harmonic Motion

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:

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 , :

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 .

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.

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:

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 .

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:

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

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.

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) .