Two particles A and B of equal masses are suspended from two massless springs of spring constants k1

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

In SHM, the displacement is given by $x = A\sin(\omega t)$. Velocity is the rate of change of displacement, so $v = \frac{dx}{dt} = A\omega\cos(\omega t)$. The maximum value of $\cos(\omega t)$ is $1$, making the maximum velocity $v_{\text{max}} = A\omega$.

Q: Does the mass of the particles affect the final ratio?

In this specific problem, no. Because the problem states that both particles have *equal* masses, the mass term $m$ cancels out when we take the ratio of their angular frequencies. If the masses were different, the final ratio would depend on both the spring constants and the masses.

Q: Where does the particle experience this maximum velocity?

A particle in SHM experiences its maximum velocity exactly as it passes through its mean (equilibrium) position. At the extreme positions (maximum amplitude), its velocity is momentarily zero.

Q: Why do we use angular frequency $\omega$ instead of regular frequency $f$?

Angular frequency $\omega$ (measured in radians per second) is mathematically more convenient when dealing with trigonometric functions in SHM equations. They are directly related by $\omega = 2\pi f$, so you could use either, but $\omega$ keeps the formulas cleaner.

Q: What if the problem asked for the ratio of their maximum accelerations?

The maximum acceleration in SHM is given by $a_{\text{max}} = A\omega^2$. You would set up a similar equation $A_1\omega_1^2 = A_2\omega_2^2$ and solve for the ratio using the same substitution for $\omega$.

Question

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_1$ and $k_2$, respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two identical masses, each of mass $m$, hanging from two different springs. The first spring has a stiffness or spring constant of $k_1$, and the second has a spring constant of $k_2$. Both of these systems are set into simple harmonic motion (SHM).

The problem gives us a very specific and crucial piece of information: the **maximum velocities** of both particles during their oscillations are exactly the same. Our goal is to find the ratio of their amplitudes, $A_1$ and $A_2$.

### The Master Equation

To solve this, we need to recall the formula for the maximum velocity of a particle executing SHM. The velocity of a particle in SHM is given by the derivative of its displacement. If displacement is $x = A\sin(\omega t)$, then velocity is $v = A\omega\cos(\omega t)$.

The maximum value of the cosine function is $1$, so the maximum velocity is simply the product of the amplitude and the angular frequency:

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

According to the problem, the maximum velocity of particle A is equal to the maximum velocity of particle B. We can write this mathematically as:

$$A_1\omega_1 = A_2\omega_2$$

By rearranging this equation, we can isolate the ratio of their amplitudes, which is what we are looking for:

$$\frac{A_1}{A_2} = \frac{\omega_2}{\omega_1}$$

### Final Calculation

Now, we need to express the angular frequencies $\omega_1$ and $\omega_2$ in terms of the given spring constants and masses. For a spring-mass system, the angular frequency is determined by the stiffness of the spring and the inertia of the mass:

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

Since both particles have the same mass $m$, we can write the angular frequencies for each system as:

$$\omega_1 = \sqrt{\frac{k_1}{m}} \quad 	ext{and} \quad \omega_2 = \sqrt{\frac{k_2}{m}}$$

Let's substitute these expressions back into our amplitude ratio equation:

$$\frac{A_1}{A_2} = \frac{\sqrt{\frac{k_2}{m}}}{\sqrt{\frac{k_1}{m}}}$$

Notice how the mass $m$ appears in the denominator of both square roots. Because the masses are equal, they beautifully cancel each other out! This leaves us with our final, elegant result:

$$\frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}}$$

This tells us that the ratio of their amplitudes is inversely proportional to the square root of their spring constants. A stiffer spring (larger $k$) will result in a smaller amplitude if the maximum velocities are to remain equal.

Animated Solution for Physics - Oscillations: Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $k_{1}$ and $k_{2}$ , respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is

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Visualized Solution

$S y s t e m S e t u p$

$M a x im u m V e l oc i t y in S H M$

$E q u a t in g V e l oc i t i es$

$R a t i o o f A m pl i t u d es$

$A n g u l a r F r e q u e n cy$

$S u b s t i t u t in g ω$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Simple Harmonic Motion

Analyzing the Setup

The Master Equation

Final Calculation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Velocity in SHM

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Amplitudes

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

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

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