Question

A particle performing S.H.M is found at its equilibrium at t = 1s and it is found to have a speed of 0.25m/s at t = 2s. If the period of oscillation is 6s. Calculate amplitude of oscillation
$\text{A}\text{. }\dfrac{3}{2\pi }m$
$\text{B}\text{. }\dfrac{3}{4\pi }m$
$\text{C}\text{. }\dfrac{6}{\pi }m$
$\text{D}\text{. }\dfrac{3}{8\pi }m$

Answer 1

Use the equation of motion of a particle under a simple harmonic motion. Find the angular frequency of the oscillations with the given value of time period. Also use the equation for the velocity of the particle. Then substitute the given values in the respective equations and find the amplitude.

Formula used:
$x=A\sin \left( \omega t+\phi \right)$
$v=A\omega \cos \left( \omega t+\phi \right)$
$T=\dfrac{2\pi }{\omega }$

Complete step by step answer:
When a particle is under simple harmonic motion, it equation of motion is given as $x=A\sin \left( \omega t+\phi \right)$ ….. (i),
where x is the displacement of the particle with respect to its mean position at time t, A is the amplitude of the oscillations, $\omega$ is the angular frequency of the oscillations and $\phi$ is the phase constant.
It is given that at time t = 1s, the particle is at the mean position (i.e. its equilibrium position). This means that at this time x = 0.
Substitute the values of x and t in equation (i).
$\Rightarrow 0=A\sin \left( \omega +\phi \right)$
We know that the amplitude A cannot be zero.
$\Rightarrow \sin \left( \omega +\phi \right)=0$
Sine of an angle is zero when the angle is a multiple of $\pi$.
$\Rightarrow \omega +\phi =n\pi$.
$\Rightarrow \phi =n\pi -\omega$ ….. (ii).
The velocity of the particle in SHM at a time t is given as $v=A\omega \cos \left( \omega t+\phi \right)$ ….. (iii).
It is given that at time t = 2s, the speed of the particle is 0.25m/s.
Substitute the values of t, $\phi$ and v in equation (iii).
$\Rightarrow 0.25=A\omega \cos \left( 2\omega +(n\pi -\omega ) \right)$
$\Rightarrow 0.25=A\omega \cos \left( n\pi +\omega \right)$
$\Rightarrow A=\dfrac{0.25}{\omega \cos \left( n\pi +\omega \right)}$.
Now, consider the term $\cos \left( n\pi +\omega \right)$.
$\cos \left( n\pi +\omega \right)=\cos \omega$, when n is an even integer.
$\cos \left( n\pi +\omega \right)=-\cos \omega$, when n is an odd integer.
Since we want the magnitude of the amplitude, we will only consider the positive value.
$\Rightarrow A=\dfrac{0.25}{\omega \cos \left( \omega \right)}$ …. (iv).
It is given that the time period of the SHM is T=6s.
And $T=\dfrac{2\pi }{\omega }$
$\Rightarrow 6=\dfrac{2\pi }{\omega }$
$\Rightarrow \omega =\dfrac{\pi }{3}{{s}^{-1}}$.
Substitute this value in equation (iv).
$\Rightarrow A=\dfrac{0.25}{\left( \dfrac{\pi }{3} \right)\cos \left( \dfrac{\pi }{3} \right)}=\dfrac{0.25}{\left( \dfrac{\pi }{3} \right)\left( \dfrac{1}{2} \right)}=\dfrac{3}{2\pi }m$.
Hence, the correct option is A.

Note:
Students may neglect the phase angle and consider the equation of motion of the particle in SHM as $x=A\sin \left( \omega t \right)$. However, when this equation, at time t = 1s, the particle will not be at x = 0.