Question

The initial position and velocity of a body moving in SHM with period $ T = 0.25{\text{ }}s $ are $ x = 5.0{\text{ }}cm $ and $ v = 218{\text{ }}cm/s $. What are the amplitude and phase difference of the motion?

Answer 1

Hint: In this solution, we will use the formula for simple harmonics motion, mainly the position and velocity of the simple harmonics motion. The phase difference of the motion is related to the position of the oscillator at the beginning of time.

Formula used: We will be using the following formula,
 $\Rightarrow x = A\sin (\omega t + \phi ) $ where $ x $ is the displacement of the oscillator, $ \omega $ is the angular velocity, $ t $ is the time, and $ \phi $ is the phase.

Complete step by step answer
We’ve been given that the initial position and velocity of a simple harmonic oscillator are $ x = 5.0{\text{ }}cm $ and $ v = 218{\text{ }}cm/s $ its time period is $ T = 0.25{\text{ }}s $. Let us start by finding the angular velocity of the oscillator.
 $\Rightarrow \omega = \dfrac{{2\pi }}{T} $.
Since $ T = 0.25{\text{ }}s $,
 $\Rightarrow \omega = 8\pi \,rad/s $
The initial position that is at $ t = 0 $ is $ x = 5.0{\text{ }}cm $. Substituting these values in $ x = A\sin (\omega t + \phi ) $, we get:
 $\Rightarrow 5 = A\sin \phi $
Now the velocity of the harmonic oscillator can be calculated using:
 $\Rightarrow v = \dfrac{{dx}}{{dt}} $
Substituting the value of $ x = A\sin (\omega t + \phi ) $ in the above equation, and taking the derivative, we can write
 $\Rightarrow v = A\omega \cos (\omega t + \phi ) $
Since the initial velocity is $ v = 218{\text{ }}cm/s $, we can substitute that at time $ t = 0 $ and write
 $\Rightarrow 218 = A\omega \cos \phi $
Taking the ratio of equation (1) and equation (2), we get
 $\Rightarrow \dfrac{5}{{218}} = \dfrac{{\tan \phi }}{{8\pi }} $
 $\Rightarrow \tan \phi = \dfrac{{5 \times 8\pi }}{{218}} $
This gives us
 $\Rightarrow \phi = 30^\circ $
Substituting $ \phi $ in equation (1), we get
 $\Rightarrow 5 = A\sin 30^\circ $
 $ \Rightarrow 5 = A \times \dfrac{1}{2} $
Therefore we get,
 $\Rightarrow A = 10\,cm $
Hence the amplitude of oscillation will be $ A = 10\,cm $ and the phase difference of the simple harmonic oscillator in question will be $ \phi = 30^\circ $.

Note
To solve this question, we must be familiar with the equations of simple harmonic motion and the explanation of different terms. The oscillator in our case starts at an arbitrary point in its motion with a non-zero position and velocity. The units of the amplitude will be the same as the units of displacement provided to us.