JEE Advanced (1981)

LEVELJEE Main

Animated Solution for Physics - Oscillations: Two masses $m_{1}$ and $m_{2}$ are suspended together by a massless spring of spring constant $k$ as shown in the figure. When the masses are in equilibrium, $m_{1}$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $m_{2}$ .

Q: Why is the initial position of $m_2$ considered the extreme position of the oscillation?

Because mass $m_1$ is removed 'without disturbing the system', meaning the remaining mass $m_2$ starts from rest at that instantaneous position. In simple harmonic motion, any position where a particle is released from rest is an extreme position.

Q: Does the angular frequency depend on the mass $m_1$ that was removed?

No. Once $m_1$ is removed, it is no longer part of the oscillating system. The only mass undergoing simple harmonic motion is $m_2$, so the angular frequency depends only on $k$ and $m_2$.

Q: What would happen to the amplitude if we removed $m_2$ instead of $m_1$?

If we removed $m_2$ instead, the remaining mass would be $m_1$. By symmetry, the new amplitude of oscillation for $m_1$ would be $A = \frac{m_2 g}{k}$.

Q: Why does the spring extension not change instantly when $m_1$ is cut off?

Due to the inertia of the remaining mass $m_2$ and the spring, physical displacements require finite time to occur. An instantaneous change in position would require infinite velocity, which is physically impossible.

Q: Is the total mechanical energy of the oscillating system conserved after $m_1$ is removed?

Yes, after $m_1$ is removed, there are no non-conservative forces acting on the remaining mass-spring system, so its mechanical energy is conserved during the subsequent oscillations.

Visualized Solution

Understanding the Initial Equilibrium State

Initially, both masses $m_{1}$ and $m_{2}$ are suspended from the spring of constant $k$ .
The system is in static equilibrium.
Let the initial extension of the spring be $x_{2}$ .

Setting up the Initial Force Balance

Total downward force: $F_{down} = (m_{1} + m_{2}) g$
Upward spring force: $F_{spring} = k x_{2}$
Equilibrium condition:
k x_2 = (m_1 + m_2)g \implies x_2 = \frac{(m_1 + m_2)g}{k}

Removing Mass $m_{1}$

Mass $m_{1}$ is suddenly removed without disturbing the system.
The remaining mass is only $m_{2}$ .
At the instant of removal, the spring is still stretched by $x_{2}$ .

Finding the New Equilibrium Position

With only $m_{2}$ left, the new equilibrium position will have a different extension, say $x_{1}$ .
New force balance:
k x_1 = m_2 g \implies x_1 = \frac{m_2 g}{k}

Determining the Angular Frequency $ω$

The system now consists of a single mass $m_{2}$ oscillating on a spring of constant $k$ .
The angular frequency of a spring-mass system is:
\omega = \sqrt{\frac{k}{m_{\text{oscillating}}}}
Since only $m_{2}$ is oscillating:
\omega = \sqrt{\frac{k}{m_2}}

Calculating the Amplitude $A$

The amplitude $A$ is the distance between the extreme position and the mean (equilibrium) position.
Extreme position extension: $x_{2}$
Mean position extension: $x_{1}$
Therefore:
A = x_2 - x_1

Substituting the Values of $x_{1}$ and $x_{2}$

Substitute $x_{2} = \frac{( m _{1} + m _{2} ) g}{k}$ and $x_{1} = \frac{m _{2} g}{k}$ into the amplitude equation:
A = \frac{(m_1 + m_2)g}{k} - \frac{m_2 g}{k}
A = \frac{m_1 g + m_2 g - m_2 g}{k} = \frac{m_1 g}{k}

Summary of Results

Angular Frequency:
\omega = \sqrt{\frac{k}{m_2}}
Amplitude of Oscillation:
A = \frac{m_1 g}{k}

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The Sigma Insight: Force and Energy Method in SHM

Solution Diagram

Analyzing the Setup

Imagine a vertical spring hanging from a rigid ceiling.

When we attach two masses,

m_{1}

and

m_{2}

, to its lower end, the spring stretches under their combined weight.

At this stage, the system is in a state of static equilibrium.

The downward gravitational force acting on the combined mass is exactly balanced by the upward restoring force of the spring.

Let's denote this initial extension of the spring as

x_{2}

Using Hooke's Law, we can write the force balance equation as:

k x_{2} = (m_{1} + m_{2}) g

From this, we find the initial extension:

x_{2} = \frac{( m _{1} + m _{2} ) g}{k}

The Sudden Change

Removing

m_{1}

Now, let's perform a thought experiment.

What happens if we suddenly remove the lower mass,

m_{1}

, without disturbing the remaining mass,

m_{2}

Because the removal is instantaneous and gentle, the spring does not immediately change its length.

Therefore, at the exact instant of removal, the remaining mass

m_{2}

is still at the position corresponding to the extension

x_{2}

However, this position is no longer the equilibrium position for

m_{2}

alone!

Since the system is released from rest at this position, it becomes the extreme position of the subsequent simple harmonic motion.

Finding the New Equilibrium Position

With only

m_{2}

suspended, the system will oscillate about a new equilibrium position.

Let's find where this new equilibrium position lies.

At this new mean position, the spring is stretched by a smaller amount,

x_{1}

, such that the spring force balances only the weight of

m_{2}

k x_{1} = m_{2} g

This gives us the new equilibrium extension:

x_{1} = \frac{m _{2} g}{k}

Calculating Angular Frequency and Amplitude

Since the oscillating system consists of only mass

m_{2}

attached to the spring of constant

k

, the angular frequency

ω

is determined solely by these two parameters:

ω = \frac{k}{m _{2}}

Next, let's find the amplitude

A

of the oscillation.

By definition, the amplitude is the distance between the extreme position and the mean position.

Since the extreme position corresponds to the extension

x_{2}

and the mean position corresponds to the extension

x_{1}

, the amplitude is simply:

A = x_{2} - x_{1}

Substituting the values of

x_{2}

and

x_{1}

we derived earlier:

A = \frac{( m _{1} + m _{2} ) g}{k} - \frac{m _{2} g}{k}

A = \frac{m _{1} g}{k}

This is a remarkably elegant result!

It shows that the amplitude of oscillation depends only on the weight of the removed mass

m_{1}

and the spring constant

k

, completely independent of the remaining mass

m_{2}

Common Traps & FAQs

Why is the initial position of

m_{2}

considered the extreme position of the oscillation?

Because mass

m_{1}

is removed 'without disturbing the system', meaning the remaining mass

m_{2}

starts from rest at that instantaneous position. In simple harmonic motion, any position where a particle is released from rest is an extreme position.

Does the angular frequency depend on the mass

m_{1}

that was removed?

No. Once

m_{1}

is removed, it is no longer part of the oscillating system. The only mass undergoing simple harmonic motion is

m_{2}

, so the angular frequency depends only on

k

and

m_{2}

What would happen to the amplitude if we removed

m_{2}

instead of

m_{1}

If we removed

m_{2}

instead, the remaining mass would be

m_{1}

. By symmetry, the new amplitude of oscillation for

m_{1}

would be

A = \frac{m _{2} g}{k}

Why does the spring extension not change instantly when

m_{1}

is cut off?

Due to the inertia of the remaining mass

m_{2}

and the spring, physical displacements require finite time to occur. An instantaneous change in position would require infinite velocity, which is physically impossible.

Is the total mechanical energy of the oscillating system conserved after

m_{1}

is removed?

Yes, after

m_{1}

is removed, there are no non-conservative forces acting on the remaining mass-spring system, so its mechanical energy is conserved during the subsequent oscillations.

Visualized Solution

Understanding the Initial Equilibrium State

Setting up the Initial Force Balance

Removing Mass $m_{1}$

Finding the New Equilibrium Position

Determining the Angular Frequency $ω$

Calculating the Amplitude $A$

Substituting the Values of $x_{1}$ and $x_{2}$

Summary of Results

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Sudden Change

Finding the New Equilibrium Position

Calculating Angular Frequency and Amplitude

Similar Questions

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless springs of spring constants $k_{1}$ and $k_{2}$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of $M$ to that of $N$ is

A mass $M$ attached to a spring oscillates with a period of $2 s$ . If the mass is increased by $2 kg$ the period increases by one sec. Find the initial mass $M$ assuming that Hooke's law is obeyed.

A point mass $m$ is suspended at the end of massless wire of length $L$ and cross-sectional area $A$ . If $Y$ is the Young's modulus of elasticity of the material of the wire, obtain the expression for the frequency of the simple harmonic motion along the vertical line.

Visualized Solution

Understanding the Initial Equilibrium State

Setting up the Initial Force Balance

Removing Mass .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​

Finding the New Equilibrium Position

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

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

Substituting the Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​

Summary of Results

The Sigma Insight: Force and Energy Method in SHM

Analyzing the Setup

The Sudden Change

Finding the New Equilibrium Position

Calculating Angular Frequency and Amplitude

Similar Questions

Removing Mass $m_{1}$

Determining the Angular Frequency $ω$

Calculating the Amplitude $A$

Substituting the Values of $x_{1}$ and $x_{2}$