Question

The amplitude of a particle in SHM is 5 cm and its time period is $\pi$. At a displacement of 3 cm from mean position the velocity in $cm{s^{ - 1}}$ will be
A) 8
B) 12
C) 2
D) 16

Answer 1

Amplitude is the maximum displacement of a particle in SHM (Simple harmonic motion) and the equation of SHM is $x = A\sin \left( {\omega t} \right)$, where A is the amplitude, x is the displacement, ω is the angular frequency, t is the time.

Complete step by step answer:
We are given that the amplitude of a particle in SHM is 5 cm and its time period is $\pi$.
We have to calculate the velocity when the displacement from mean position is 3cm.
We know that, the equation of SHM is
$x = A\sin \left( {\omega t} \right)$
Velocity can be defined as the product of the Amplitude and the angular frequency.
$\therefore V = A\omega \cos \left( {\omega t} \right)$
We also know that time period is inversely proportional to the angular frequency.
$T = \dfrac{{2\pi }}{\omega }$
We are already given that the time period is $\pi$
$T = \pi \\\ \Rightarrow T = \dfrac{{2\pi }}{\omega } \\\ \Rightarrow \pi = \dfrac{{2\pi }}{\omega } \\\ \Rightarrow \omega = \dfrac{{2\pi }}{\pi } \\\ \Rightarrow \omega = 2 \\\$
Substituting the value of angular frequency and amplitude in $x = A\sin \left( {\omega t} \right)$
$A = 5cm,\omega = 2 \\\ x = 5\sin \left( {2t} \right) \\\ \Rightarrow x = 5\sin 2t \\\$
At displacement x=3 cm,
$\Rightarrow x = 3 \\\ \Rightarrow 3 = 5\sin 2t \\\ \Rightarrow \sin 2t = \dfrac{3}{5} \\\ \Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta } \\\ \Rightarrow \cos 2t = \sqrt {1 - {{\sin }^2}2t} \\\ \Rightarrow \cos 2t = \sqrt {1 - \dfrac{9}{{25}}} \\\ \Rightarrow \cos 2t = \sqrt {\dfrac{{16}}{{25}}} \\\ \Rightarrow \cos 2t = \dfrac{4}{5} \\\$
Velocity V is $V = A\omega \cos \left( {\omega t} \right)$, substitute the values of A, cos2t, ω to calculate velocity.
$V = A\omega \cos \left( {\omega t} \right) \\\ A = 5cm,\omega = 2,\cos 2t = \dfrac{4}{5} \\\ \Rightarrow V = 5 \times 2 \times \dfrac{4}{5} \\\ \Rightarrow V = \dfrac{{40}}{5} \\\ \therefore V = 8cm{s^{ - 1}} \\\$
Velocity when the displacement of the particle from mean position is 3cm is 8cm/s.

The correct option is option A , 8.

Note: Velocity of the object in SHM is maximum at the mean position. We have cosine function in the velocity and sine function in the displacement formula because velocity is the rate of change of displacement and we have to differentiate displacement with respect to time to find velocity. And the differentiation of Sine is cosine so we have used cosine function for velocity.