Question

The motion of a particle executing SHM is given $x=0.01\sin \pi (t+0.05)$, where x is in meters and t is in second. The time period of motion (in second) is
A. 0.01
B. 0.02
C. 0.1
D. 0.2

Answer 1

Hint:We have given the motion of the particle which executes SHM and we have to find the time period of motion. First we compare the given equation with the standard formula of simple harmonic motion, then by using the formula of angular frequency and putting the values, we find the time period and choose the correct option.

Formula Used:
Formula for the amplitude of the simple harmonic motion is
$x=A\sin (wt+\phi )$
And $w=\dfrac{2\pi }{T}$
Where x is the displacement of the particle, W is the angular frequency, A is its amplitude, T is the time taken for its simple harmonic motion and T is the time taken for the particular position.

Complete step by step solution:
Given that $x=0.01\sin \pi (t+0.05)$…………………… (1)
we use the formula of SHM
$x=A\sin (wt+\phi )$
We know $w=\dfrac{2\pi }{T}$
Then $x=A\sin (\dfrac{2\pi }{T}t+\phi )$………………………………… (2)
On comparing the equation (1) with equation (2), we get
$A=0.01$, $\phi =0.05$
And,
$w=100\pi $
We know,
$w=\dfrac{2\pi }{T}$
Therefore,
$T=\dfrac{2\pi }{w}$
Then,
$T=\dfrac{2\pi }{100\pi }$= $\dfrac{1}{50}s \\ $
$\therefore T=0.02\,s$

Thus, option B is the correct answer.

Note: In this question, we use the formula $x=A\sin (wt+\phi )$. This is because there exists a phase of an oscillating particle regarding its displacement and direction of vibration at that particular instant. $(wt+\phi )$ is the phase of the particle in which the particle is oscillating. At time t = 0 the phase angle is known as the initial phase.