Question

What is the phase difference between two harmonic motions represented by $x_1 = A\sin (\omega t + \dfrac{\pi }{6})$ and $x_2 = A\cos (\omega t)$ .
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{2}$
D. $\dfrac{{2\pi }}{3}$

Answer 1

We can define the phase of a particle in simple harmonic motion as the initial stage of Simple harmonic motion which is described as an angular term that indicates a particle's state relative to its mean position at a given instant. To calculate the phase difference of two particles following simple harmonic motion we need to first convert the equation in the same trigonometric form and then find the difference in phase.

Complete Step By Step Answer:
The first equation is $x_1 = A\sin (\omega t + \dfrac{\pi }{6})$
The second equation is $x_2 = A\cos (\omega t)$
Now writing them in the same format,
For the first one $x_1 = A\sin (\omega t + \dfrac{\pi }{6})$ remains in $\sin$ form.
Now, the second one $x_2 = A\cos (\omega t) = A\sin (\dfrac{\pi }{2} + \omega t)$ .
The phase is called $(\omega t + \phi )$ , where $\phi$ is called the initial phase.
So, the difference is
$(\omega t + \dfrac{\pi }{2}) - (\omega t + \dfrac{\pi }{6}) \\\ = \dfrac{\pi }{3}$
So, by the above calculation we saw that to calculate the phase difference between two particles following simple harmonic motion we need to get them in the same trigonometric form or any other mathematical form. After that we just have to get their differences.
Now, we can see that according to the calculation done above the correct option is B.

Note:
Simple harmonic motion is a type of periodic motion in mechanics and physics in which the restoring force on a moving object is directly proportional to the size of the object's displacement and acts in the direction of the object's equilibrium position. For example, Simple pendulum swings back and forth in the same line when we swing it. Oscillations are what these movements are. Simple harmonic motion here is demonstrated by the oscillations of a pendulum.