Two bodies M and N of equal masses are suspended from two separate massless springs of spring consta

Q: Why is the maximum velocity equal to $\omega A$?

The displacement in SHM is given by $x(t) = A \sin(\omega t + \phi)$. Differentiating this with respect to time gives the velocity $v(t) = A\omega \cos(\omega t + \phi)$. The maximum value of velocity occurs when the cosine term is at its maximum value of $1$, which gives $v_{\text{max}} = \omega A$. This occurs as the particle passes through the equilibrium position.

Q: Does the mass of the blocks affect the ratio of the amplitudes?

In this specific problem, no. Because the problem states that the two bodies have equal masses, the mass term $m$ cancels out during the algebraic simplification. However, if the masses were different, the ratio of amplitudes would depend on both the spring constants and the masses: $\frac{A_M}{A_N} = \sqrt{\frac{k_2}{k_1} \cdot \frac{m_M}{m_N}}$.

Q: What is the physical meaning of a larger spring constant $k$?

A larger spring constant $k$ represents a stiffer spring. It requires more force to stretch or compress by a unit distance. In an oscillatory system, a stiffer spring provides a larger restoring force, leading to higher acceleration and a faster rate of oscillation (higher angular frequency $\omega$).

Q: Why is the amplitude ratio inversely proportional to the square root of the spring constants?

For equal maximum velocities, the product $\omega A$ must be constant for both systems. Since angular frequency $\omega$ is directly proportional to the square root of the spring constant ($\omega \propto \sqrt{k}$), the amplitude $A$ must be inversely proportional to the square root of the spring constant ($A \propto 1/\sqrt{k}$) to keep the product constant.

Q: Where does the maximum velocity occur during the oscillation?

The maximum velocity always occurs at the mean (equilibrium) position ($x = 0$). At this point, the potential energy of the spring is zero, and all the energy of the system is in the form of kinetic energy.

Question

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

Options:

(A) $\frac{k_1}{k_2}$

(B) $\sqrt{\frac{k_2}{k_1}}$

(C) $\frac{k_2}{k_1}$

(D) $\sqrt{\frac{k_1}{k_2}}$

Sigma Percentile · Accepted Answer

### Introduction to Simple Harmonic Motion

Simple Harmonic Motion (SHM) is one of the most fundamental and elegant concepts in physics. It describes the back-and-forth oscillatory motion of a system governed by a restoring force that is directly proportional to the displacement from its equilibrium position.

In this problem, we explore the relationship between the physical parameters of two independent spring-mass systems. Specifically, we are asked to find the ratio of their amplitudes given that their maximum velocities are equal. This problem beautifully highlights how stiffness (represented by the spring constant $k$) and inertia (represented by the mass $m$) dictate the kinematics of oscillation.

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### Deconstructing the Setup

Imagine two identical blocks, each of mass $m$, suspended from two separate vertical springs.

* The first spring has a spring constant of $k_1$.
* The second spring has a spring constant of $k_2$.

When these blocks are displaced from their equilibrium positions and released, they begin to oscillate vertically. The motion of each block is simple harmonic.

Let the amplitude of oscillation for the first block (Body $M$) be $A_M$, and for the second block (Body $N$) be $A_N$.

Our goal is to find the ratio:

$$\frac{A_M}{A_N}$$

under the constraint that their maximum velocities are equal.

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### The Physics of Speed in SHM

To solve this, we must recall the kinematic equations of SHM. The displacement $x(t)$ of a particle executing SHM can be written as:

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

where $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the phase constant. Differentiating this with respect to time $t$ gives the velocity $v(t)$:

$$v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$$

Since the maximum value of the cosine function is $1$, the **maximum velocity** $v_{	ext{max}}$ occurs when the block passes through the mean position ($x = 0$):

$$v_{	ext{max}} = \omega A$$

This equation is our primary tool. It tells us that the maximum speed is directly proportional to both the angular frequency and the amplitude of oscillation.

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### Connecting the Systems

We are given that the maximum velocities of both bodies are equal:

$$(v_M)_{	ext{max}} = (v_N)_{	ext{max}}$$

Substituting our expression for maximum velocity into this equality, we get:

$$\omega_M A_M = \omega_N A_N$$

Rearranging this equation to solve for the ratio of the amplitudes yields:

$$\frac{A_M}{A_N} = \frac{\omega_N}{\omega_M}$$

This is a crucial conceptual milestone. It shows that for two systems to achieve the same maximum speed, their amplitudes must be **inversely proportional** to their angular frequencies.

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### The Elegant Simplification

Now, we need to express the angular frequencies in terms of the given physical properties of the systems. For a spring-mass system, the angular frequency is determined by the stiffness of the spring and the mass of the block:

$$\omega = \sqrt{\frac{k}{m}}$$

Since the masses of both bodies are equal ($m_M = m_N = m$), we can write the angular frequencies of the two systems as:

$$\omega_M = \sqrt{\frac{k_1}{m}}$$

$$\omega_N = \sqrt{\frac{k_2}{m}}$$

Substituting these expressions back into our amplitude ratio equation:

$$\frac{A_M}{A_N} = \frac{\sqrt{\frac{k_2}{m}}}{\sqrt{\frac{k_1}{m}}}$$

Notice how the mass term $m$ is present in both the numerator and the denominator under the square root. Because the masses are identical, they cancel out completely:

$$\frac{A_M}{A_N} = \sqrt{\frac{k_2}{k_1}}$$

This is our final, elegant result. It matches **Option (b)** perfectly.

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### Intuitive Check: Why the Inverse Square Root?

Let's think about the physical intuition behind this result.

A stiffer spring (larger $k$) exerts a stronger restoring force for any given displacement. This larger force causes the mass to accelerate more rapidly, leading to a higher angular frequency $\omega$.

If a system is oscillating with a very high frequency, it completes its cycles much faster. To keep its maximum velocity the same as a slower system, it does not need to travel as far from the equilibrium position. Therefore, the stiffer system must have a **smaller amplitude** of oscillation.

Mathematically, since $A \propto \frac{1}{\omega}$ for a constant $v_{	ext{max}}$, and $\omega \propto \sqrt{k}$, it naturally follows that:

$$A \propto \frac{1}{\sqrt{k}}$$

This explains why the ratio of the amplitudes is proportional to the square root of the inverse ratio of their spring constants.

Select Answer:

Visualized Solution

Visualize the Two Spring-Mass Systems

Recall SHM Formulas

Equating Maximum Velocities

Expressing the Amplitude Ratio

Substituting Angular Frequencies

Simplifying the Ratio

Physical Intuition & Variations

The Sigma Insight: Force and Energy Method in SHM

Introduction to Simple Harmonic Motion

Deconstructing the Setup

The Physics of Speed in SHM

Connecting the Systems

The Elegant Simplification

Intuitive Check

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