Question

An elastic string of length $l$ supports a heavy particle of mass m and the system is in equilibrium with elongation produced being e as shown in figure. The particle is now pulled down below the equilibrium position through a distance $d( \leqslant e)$ and released. The angular frequency and maximum amplitude for SHM is
A. $\sqrt {\dfrac{g}{e}} ,e$
B. $\sqrt {\dfrac{g}{l}} ,2e$
C. $\sqrt {\dfrac{g}{{d + e}}} ,d$
D. $\sqrt {\dfrac{g}{e}} ,2d$

Answer 1

Hint: System is in equilibrium means, the applied force is being opposed by the equal and opposite force. And hence, the net force applied on the body is zero. When the particle is pulled down from the equilibrium position, a restoring force will be applied on it which will try to pull it back to the equilibrium position leading to simple harmonic motion of the body.

Formula used:
Where,
$kx$ is the restoring force.
$mg$ is the force due to acceleration due to gravity.
$T = 2\pi \sqrt {\dfrac{m}{k}} $
Where,
$T$ is time period
$m$ is mass

Complete step by step answer:
It is given that, the length of the string is $l$
And the mass of the particle is $m$.
Now, the distance covered by the particle after pulling it down is $d$.
Since, the system is in equilibrium. Weight of the ball must be balanced by the spring force.
$ \Rightarrow kx = mg$
Where,
$x$ is the distance by which a body is pulled against the restoring force
And $k$is the restoring constant.
$ \Rightarrow ke = mg$
Where, $e$ is the length of equilibrium position.
$ \Rightarrow k = \dfrac{{mg}}{e}$
When the particle is pulled down by distance $d$ the force will be
$F = kd$
We also know that
$F = ma,$
Therefore, we get
$ma = kd$
By rearranging it, we get
$ \Rightarrow a = \dfrac{{kd}}{m}$ . . . (1)
We know that,
$a = {\omega ^2}d$
Where,
$a$ is linear acceleration
$\omega $ is angular frequency.
By substituting this value in equation (1), we get
\[{\omega ^2}d = \dfrac{{kd}}{m}\]
By cancelling the common terms, we get
${\omega ^2} = \dfrac{k}{m}$
$ \Rightarrow \omega = \sqrt {\dfrac{{\dfrac{{mg}}{e}}}{m}} $ $\left( {\because k = \dfrac{{mg}}{e}} \right)$
By cancelling the common term and re-arranging it, we get
$ \Rightarrow \omega = \sqrt {\dfrac{g}{e}} $
We know that,
$T = \dfrac{{2\pi }}{\omega }$
Where,
$T$ is time period
By substituting the value of $\omega $ in the above equation, we get
$ \Rightarrow T = \dfrac{{2\pi }}{{\sqrt {\dfrac{g}{e}} }}$
$ \Rightarrow T = 2\pi \sqrt {\dfrac{e}{g}} $
Now, at extreme position,
$g = {\omega ^2}A$ $\left( {\because a = g} \right)$
Where,
$A$ is the distance from center to the point at extreme position.
$ \Rightarrow g = {\left( {\sqrt {\dfrac{g}{e}} } \right)^2} \times A$ \[\left( {\because \omega = \sqrt {\dfrac{g}{e}} } \right)\]
$ \Rightarrow g = \dfrac{g}{e} \times A$
By cancelling the common term, we get
$ \Rightarrow 1 = \dfrac{A}{e}$
$ \Rightarrow e = A$
Therefore, the angular frequency is,
$\omega = \sqrt {\dfrac{g}{e}} $
And maximum amplitude is $A = e$

Therefore, from the above explanation, the correct answer is, option is (A), $\sqrt {\dfrac{g}{e}} ,e$.

So, the correct answer is “Option A”.

Note: You need to understand how the concept of equilibrium forces were to tackle such types of questions in the future. Restoring force is the force that tries to get the particle back to the equilibrium position. Restoring force is a variable force. It keeps on changing according to the real force applied on the particle.