Question

Two bodies $M$ and $N$ of equal masses are suspended from two separate massless spring of spring constants ${k_1}$ and ${k_2}$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, then the ratio of amplitude of $M$ to that of $N$ is:
A) $\dfrac{{{k_2}}}{{{k_1}}}$
B) $\sqrt {\dfrac{{{k_2}}}{{{k_1}}}} $
C) $\dfrac{{{k_1}}}{{{k_2}}}$
D) $\sqrt {\dfrac{{{k_1}}}{{{k_2}}}} $

Answer 1

Hint: Recall that spring is a device that stores potential energy and works on the principle of Hooke’s law. Remember that an oscillating spring performs simple harmonic motion, so to find the ratio of amplitude we can use the equation of velocity of SHM.

Complete step by step solution:
An oscillating spring performs simple harmonic motion where simple harmonic motion is a periodic to and fro motion such that the maximum displacement on both sides of the mean position is always equal.
Mathematically, the equation of displacement of simple harmonic equation is given by
$x = A\cos (\omega t)$
Where $x$ is the displacement
$A$ is the amplitude
$\omega $ is the frequency ($\omega = \sqrt {\dfrac{k}{m}} $ where $k$ is the spring constant and $m$ is the mass) , $t$ is the time
Differentiating this equation, we get the equation for the velocity as
$v = \dfrac{{dx}}{{dt}} = A\omega \sin (\omega t)$
By the equation we know that the maximum velocity will be when $\sin (\omega t) = 1$ i.e. the maximum velocity is $A\omega $
$ \Rightarrow {v_{\max }} = A\omega $
Now that we know the equation of maximum velocity, we can find the ratio of the amplitude in terms of spring constant by substituting the frequency in terms of spring constant.
$ \Rightarrow {v_1} = {A_1}{\omega _1}$ ---( for body $M$ )
$ \Rightarrow {v_2} = {A_2}{\omega _2}$ ---( for body $N$ )
The question states that the maximum velocities for both the bodies are equal.
$ \Rightarrow {v_1} = {v_2}$
$ \Rightarrow {A_1}{\omega _1} = {A_2}{\omega _2}$
$ \Rightarrow {A_1}\sqrt {\dfrac{{{k_1}}}{m}} = {A_2}\sqrt {\dfrac{{{k_2}}}{m}} $ ---( substituting $\omega = \sqrt {\dfrac{k}{m}} $ )
Rearranging the terms, we get
$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \dfrac{{\sqrt {\dfrac{{{k_2}}}{m}} }}{{\sqrt {\dfrac{{{k_1}}}{m}} }}$
$ \Rightarrow \dfrac{{{A_1}}}{{{A_2}}} = \sqrt {\dfrac{{{k_2}}}{{{k_1}}}} $
Thus, we now know that the ratio of the amplitudes of the bodies is equal to the square root of the reciprocal of their spring constants.

Therefore, option (B), $\dfrac{{{A_1}}}{{{A_2}}} = \sqrt {\dfrac{{{k_2}}}{{{k_1}}}} $ is the correct answer.

Note: A body performs simple harmonic motion if it satisfies two conditions: (i) The acceleration of the body is always in the opposite direction to that of the displacement from the equilibrium position. (ii) The acceleration of the body should be proportional to the displacement of the body from the mean position.