A block of mass m attached to a massless spring is performing oscillatory motion of amplitude A on a

Q: Why does the velocity remain the same when the mass breaks off?

The mass breaks off gently without any external impulsive force acting along the direction of motion. By Newton's first law, the remaining mass continues with the same velocity it had at that exact instant.

Q: Why did we use $E = \frac{1}{2} k A^2$ instead of $\frac{1}{2} m v^2$?

Both are perfectly equal at the equilibrium position! Using $\frac{1}{2} k A^2$ makes it much easier to compare the old amplitude $A$ with the new amplitude $fA$ because the spring constant $k$ remains unchanged.

Q: Does the time period of the oscillation change?

Yes! The time period is $T = 2\pi \sqrt{\frac{m}{k}}$. Since the mass becomes half, the new time period becomes $T' = 2\pi \sqrt{\frac{m/2}{k}} = \frac{T}{\sqrt{2}}$. The system actually oscillates faster!

Q: What would happen if the mass was added instead of removed?

If mass was gently dropped onto the block at equilibrium, it would be an perfectly inelastic collision. The velocity would decrease to conserve linear momentum, and the amplitude would decrease as a result.

Q: What if the mass broke off at the extreme position?

At the extreme position, the velocity is zero, and all energy is potential energy ($\frac{1}{2} k A^2$). Breaking off mass doesn't change the stretch of the spring, so the potential energy and the amplitude would remain exactly the same!

Question

A block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude $A$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system becomes $fA$. The value of $f$ is

Options:

(A) $\sqrt{2}$

(B) $\frac{1}{2}$

(C) $\frac{1}{\sqrt{2}}$

(D) $1$

Sigma Percentile · Accepted Answer

### The Setup: Maximum Velocity at Equilibrium

Imagine a block of mass $m$ oscillating back and forth on a frictionless horizontal surface, attached to a massless spring with a spring constant $k$. As the block passes through its equilibrium position ($x=0$), it reaches its maximum speed, $v_{	ext{max}}$.

At this exact instant, the spring is neither stretched nor compressed. This means that the potential energy of the system is zero, and **all the mechanical energy of the system is purely kinetic**.

We can express this total energy $E$ in two equivalent ways:
$$E = \frac{1}{2} m v_{	ext{max}}^2 = \frac{1}{2} k A^2$$

### The Event: A Sudden Loss of Mass

Now, right as the block crosses the equilibrium point, a dramatic event occurs: **half of the mass suddenly breaks off!**

Because the mass breaks off gently without any external impulsive force acting along the direction of motion, the remaining half of the block continues to move forward with the exact same maximum velocity, $v_{	ext{max}}$.

However, because the moving mass is now halved ($m' = m/2$), the kinetic energy of our new system is also exactly halved.

$$E' = \frac{1}{2} \left(\frac{m}{2}ight) v_{	ext{max}}^2 = \frac{E}{2}$$

### The Math: Equating the Energies

Let's analyze the remaining system. The spring constant $k$ hasn't changed, but the mass is now $m/2$. The new system will oscillate with a new amplitude, which the problem defines as $fA$.

We know that the total energy of any spring-mass system can always be written as $\frac{1}{2} k 	imes (	ext{Amplitude})^2$. For our new system, this is:

$$E' = \frac{1}{2} k (fA)^2$$

We also established that this new total energy is exactly half of the original total energy:

$$E' = \frac{1}{2} \left( \frac{1}{2} k A^2 ight) = \frac{1}{4} k A^2$$

### The Grand Finale: Solving for the New Amplitude

Now, we simply equate the two expressions for the new total energy $E'$:

$$\frac{1}{2} k (fA)^2 = \frac{1}{4} k A^2$$

$$\frac{1}{2} k f^2 A^2 = \frac{1}{4} k A^2$$

Notice how beautifully the physics simplifies the math. The spring constant $k$ and the original amplitude squared $A^2$ cancel out from both sides.

$$f^2 = \frac{1}{2}$$

Taking the square root of both sides, we arrive at our final answer:

$$f = \frac{1}{\sqrt{2}}$$

This means the new amplitude is $A/\sqrt{2}$. The system lost half its energy, but because energy scales with the *square* of the amplitude, the amplitude only decreased by a factor of $\sqrt{2}$.

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Visualized Solution

$Initial Kinetic Energy$

$Mass Breaks Off$

$New Angular Frequency$

$Finding New Amplitude$

$Solving for f$

$What if it broke at extreme?$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Setup

The Event

The Math

The Grand Finale

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass Breaks Off

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Angular Frequency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding New Amplitude

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for f

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it broke at extreme?