Question

A particle of mass $0.1kg$ executes SHM under a force $F = ( - 10x)$N. Speed of particle at mean position is $6m/s.$Then amplitude of oscillations is
(A) $0.6m$
(B) $0.2m$
(C) $0.4m$
(D) $0.1m$

Answer 1

Hint Force is given as mass times acceleration. The formula of acceleration can be found which gives us the angular velocity of the particle. It is known at the mean position the velocity of the oscillating particle is maximum and is equal to angular velocity times the amplitude. This gives us the amplitude of oscillation of the particle.
Formula used:
${v_{\max }} = A\omega $
Where ${v_{\max }}$is the maximum velocity of a particle oscillating in SHM.
$A$is the acceleration of the particle.
$\omega $ is the angular velocity of the particle.

Complete Step by step solution
It is given that the force causing the simple harmonic motion is $F = ( - 10x)$ N
The mass of the particles is $m = 0.1kg$
From Newton’s second law of motion,
We know that-
$F = ma$
Where $a$is the acceleration produced in the body.
On substituting the given values,
$ - 10x = 0.1a$
$a = - 100x$
It is known that for a simple harmonic motion,
$a = - {\omega ^2}x$
Comparing the value of acceleration (a) in both cases, we have-
$a = - {\omega ^2}x = - 100x$
On solving this relation we get,
$\omega = 10rad/\operatorname{s} $
We know that the velocity of the particle $v$is given by,
$v = - \omega A\sin (\omega t + \phi )$
Where $A$is the amplitude of the oscillation.
$t$is the instantaneous time.
$\phi $is the phase difference.
In this question, there is no mention of the phase difference, therefore,
$\phi = 0$
$\omega t$ gives the value of angle covered by the oscillating particle.
At the mean position, the angle completed by the particle is equal to $\dfrac{\pi }{2}$ radian or $90^\circ $.
Therefore,
$\sin \dfrac{\pi }{2} = 1$
We also know that at the mean position, the velocity of the particle is maximum.
Therefore,
$\left| {{v_{\max }}} \right| = \left| { - A\omega } \right|$
From the question, ${v_{\max }} = 6m/s$
Thus,
$A = \dfrac{{{v_{\max }}}}{\omega }$
$A = \dfrac{6}{{10}} = 0.6$
Therefore, the amplitude of the oscillation of the particle is $0.6m$.

Option (A) is correct.

Note During its full time period, the particle executing SHM completes an angle of $2\pi $radians or $360^\circ $. This involves its movement from mean position to extreme right $(90^\circ )$, extreme right to extreme left $\left( {180^\circ } \right)$ and from extreme left to the mean position $(90^\circ )$.