A particle of mass m is attached to one end of a massless spring of force constant k, lying on a fri

Q: Why does the amplitude of the SHM not change after the collision with the wall?

Because the collision is perfectly elastic and the wall is rigid, there is no loss of kinetic energy during the impact. Since the total mechanical energy of the system is conserved, the amplitude $A$ (which is directly related to the total energy by $E = \frac{1}{2}kA^2$) remains completely unchanged.

Q: How did we get the time to reach the wall as $T/6$?

The position of the block starting from equilibrium is $x(t) = A \sin(\omega t)$. We found the wall is at $x = \frac{\sqrt{3}}{2}A$. Setting these equal gives $\sin(\omega t) = \frac{\sqrt{3}}{2}$, which yields $\omega t = \frac{\pi}{3}$. Since $\omega = \frac{2\pi}{T}$, we get $t = \frac{T}{6}$.

Q: Why is the return time from the wall to equilibrium also $T/6$?

By symmetry, the equations of motion for a conservative system are time-reversible. The path from the wall back to equilibrium is identical in distance and force profile to the path from equilibrium to the wall, so it takes the exact same amount of time.

Q: Why does the speed of the block return to $u_0$ at equilibrium?

At the equilibrium position ($x=0$), the spring is unstretched, so the potential energy is zero. By conservation of energy, all the mechanical energy of the system is kinetic. Since the total energy was conserved during the elastic collision, the kinetic energy (and thus the speed) at equilibrium must return to its initial value, $u_0$.

Q: What would happen if the collision was inelastic?

If the collision were inelastic, some kinetic energy would be lost as heat. The block would bounce back with a speed less than $0.5 u_0$, resulting in a new, smaller amplitude of oscillation for the subsequent motion, and the times for the subsequent phases would change.

Question

A particle of mass $m$ is attached to one end of a massless spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t = 0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$, it collides elastically with a rigid wall. After this collision
(2013 Adv.)

Options:

(A) the speed of the particle when it returns to its equilibrium position is $u_0$

(B) the time at which the particle passes through the equilibrium position for the first time is $t = \pi \sqrt{\frac{m}{k}}$

(C) the time at which the maximum compression of the spring occurs is $t = \frac{4\pi}{3} \sqrt{\frac{m}{k}}$

(D) the time at which the particle passes through the equilibrium position for the second time is $t = \frac{5\pi}{3} \sqrt{\frac{m}{k}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a classic spring-mass system resting on a completely frictionless horizontal table.

At $t = 0$, the block of mass $m$ is kicked from its equilibrium position ($x = 0$) with an initial velocity $u_0$ directed towards the right.

Under normal circumstances, this block would execute standard simple harmonic motion (SHM) with an amplitude $A$ and angular frequency $\omega = \sqrt{\frac{k}{m}}$.

However, a rigid wall is placed to the right of the equilibrium position, which will intercept the block before it can reach its natural right extreme position.

Let's trace the motion step-by-step and understand how this elastic collision alters the timeline of the oscillation.

### Finding the Position of the Wall

For any particle executing SHM, the relationship between its velocity $v$ and displacement $x$ is given by the well-known formula:

$$v = \omega \sqrt{A^2 - x^2}$$

At the equilibrium position ($x = 0$), the velocity is maximum and is given as $u_0$. Therefore, we have:

$$u_0 = \omega A$$

The problem states that the block collides with the wall when its speed drops to $0.5 u_0$. Let's substitute this condition into our velocity equation to find the position of the wall ($x_{	ext{wall}}$):

$$0.5 u_0 = \omega \sqrt{A^2 - x_{	ext{wall}}^2}$$

$$0.5 \omega A = \omega \sqrt{A^2 - x_{	ext{wall}}^2}$$

Squaring both sides of the equation:

$$0.25 A^2 = A^2 - x_{	ext{wall}}^2$$

$$x_{	ext{wall}}^2 = 0.75 A^2 \implies x_{	ext{wall}} = \frac{\sqrt{3}}{2}A$$

Thus, the rigid wall is located at a distance of $\frac{\sqrt{3}}{2}A$ to the right of the equilibrium position.

### Calculating the Time to Collision

Since the block starts from the equilibrium position at $t = 0$ and moves to the right, its displacement as a function of time is described by:

$$x(t) = A \sin(\omega t)$$

Let $t_1$ be the time taken by the block to reach the wall. At this instant, $x(t_1) = x_{	ext{wall}}$:

$$A \sin(\omega t_1) = \frac{\sqrt{3}}{2}A$$

$$\sin(\omega t_1) = \frac{\sqrt{3}}{2}$$

$$\omega t_1 = \frac{\pi}{3} \implies t_1 = \frac{\pi}{3\omega}$$

Since the time period of a complete oscillation is $T = \frac{2\pi}{\omega}$, we can express $t_1$ in terms of $T$:

$$t_1 = \frac{T}{6}$$

So, it takes exactly one-sixth of a time period for the block to reach the wall.

### The Elastic Collision and First Return to Equilibrium

When the block hits the rigid wall, it undergoes a perfectly elastic collision.

Because the wall is rigid and infinitely massive compared to the block, the block bounces back with the same speed but in the opposite direction.

Immediately after the collision, the velocity of the block is $-0.5 u_0$ (moving to the left).

Since the collision is perfectly elastic, no mechanical energy is lost. The total energy of the system remains conserved, meaning the amplitude $A$ and angular frequency $\omega$ of the subsequent SHM remain completely unchanged.

By symmetry, the time taken by the block to travel back from the wall ($x = \frac{\sqrt{3}}{2}A$) to the equilibrium position ($x = 0$) is equal to the time taken to reach the wall from equilibrium:

$$t_{	ext{return}} = t_1 = \frac{T}{6}$$

Therefore, the total time $t_{	ext{first}}$ at which the block passes through the equilibrium position for the first time is:

$$t_{	ext{first}} = t_1 + t_{	ext{return}} = \frac{T}{6} + \frac{T}{6} = \frac{T}{3}$$

Substituting $T = 2\pi \sqrt{\frac{m}{k}}$:

$$t_{	ext{first}} = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$$

This shows that **Option (b)** is incorrect.

Furthermore, since energy is conserved, when the block returns to the equilibrium position ($x = 0$), the potential energy of the spring is zero, and all energy is kinetic. Thus, its speed must be equal to the initial speed $u_0$. This confirms that **Option (a)** is correct.

### Reaching Maximum Compression

After passing through the equilibrium position, the block continues to move to the left towards the maximum compression point ($x = -A$).

Since there is no wall on the left side, the block will complete this part of the oscillation naturally.

The time taken to travel from the equilibrium position to the extreme position is one-quarter of a full time period:

$$t_{	ext{eq} 	o 	ext{ext}} = \frac{T}{4}$$

Therefore, the time $t_{	ext{comp}}$ at which maximum compression of the spring occurs is:

$$t_{	ext{comp}} = t_{	ext{first}} + \frac{T}{4} = \frac{T}{3} + \frac{T}{4} = \frac{7T}{12}$$

Substituting $T = 2\pi \sqrt{\frac{m}{k}}$:

$$t_{	ext{comp}} = \frac{7\pi}{6} \sqrt{\frac{m}{k}}$$

This shows that **Option (c)** is incorrect.

### Second Passage Through Equilibrium

After reaching the maximum compression point, the block momentarily stops and then begins to move back to the right, heading towards the equilibrium position once again.

The time taken to travel from the left extreme back to the equilibrium position is another quarter of a time period:

$$t_{	ext{ext} 	o 	ext{eq}} = \frac{T}{4}$$

Therefore, the total time $t_{	ext{second}}$ at which the block passes through the equilibrium position for the second time is:

$$t_{	ext{second}} = t_{	ext{comp}} + \frac{T}{4} = \frac{7T}{12} + \frac{T}{4} = \frac{5T}{6}$$

Substituting $T = 2\pi \sqrt{\frac{m}{k}}$:

$$t_{	ext{second}} = \frac{5\pi}{3} \sqrt{\frac{m}{k}}$$

This perfectly matches **Option (d)**, confirming it is correct.

### Summary of Results

* **Option (a)** is **correct** because the elastic collision conserves energy, ensuring the speed at equilibrium remains $u_0$.
* **Option (b)** is **incorrect** because the first passage time is $\frac{2\pi}{3} \sqrt{\frac{m}{k}}$, not $\pi \sqrt{\frac{m}{k}}$.
* **Option (c)** is **incorrect** because maximum compression occurs at $t = \frac{7\pi}{6} \sqrt{\frac{m}{k}}$, not $\frac{4\pi}{3} \sqrt{\frac{m}{k}}$.
* **Option (d)** is **correct** because the second passage time is indeed $\frac{5\pi}{3} \sqrt{\frac{m}{k}}$.

Select Answer:

Visualized Solution

Understanding the Physical Setup

Relating Velocity and Position in SHM

Locating the Rigid Wall

Calculating the Time to Collision

Analyzing the Elastic Collision

First Return to Equilibrium

Speed at Equilibrium

Journey to Maximum Compression

Second Passage Through Equilibrium

Conclusion and Correct Options

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

Finding the Position of the Wall

Calculating the Time to Collision

The Elastic Collision and First Return to Equilibrium

Reaching Maximum Compression

Second Passage Through Equilibrium

Summary of Results

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Select Answer:

Visualized Solution

Understanding the Physical Setup

Relating Velocity and Position in SHM

Locating the Rigid Wall

Calculating the Time to Collision

Analyzing the Elastic Collision

First Return to Equilibrium

Speed at Equilibrium

Journey to Maximum Compression

Second Passage Through Equilibrium

Conclusion and Correct Options

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Analyzing the Setup

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The Elastic Collision and First Return to Equilibrium

Reaching Maximum Compression

Second Passage Through Equilibrium

Summary of Results

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