Question

Two simple harmonic motions are represented by the equations:
${{Y}_{1}}=0.1sin\left( 100\pi t+\dfrac{\pi }{3} \right)$
${{Y}_{2}}=0.1cos\left( \pi t \right)$
The phase difference between velocity of particle 1 with respect to the velocity of particle 2 at t=0 is:
$\text{A} . \quad -\dfrac{\pi}{6}$
$\text{B} . \quad \dfrac{\pi}{6}$
$\text{C} . \quad -\dfrac{\pi}{3}$
$\text{D} . \quad \dfrac{\pi}{3}$

Answer 1

The phase difference between velocities of particles can be calculated only if we have the equations of velocities of both the particles. As the above equations represent S.H.M, so the velocities must also be of the form $v=a\omega sin\left( \omega t+\phi \right)$. Here we will use the concept that velocity is the rate of change of position.

Formula used:
$|v|=\dfrac{dx}{dt}$ i.e. magnitude to velocity (also called its speed) is the rate of change of its position.

Complete step-by-step answer:
First, let’s understand the standard wave equation.
$Y=asin\left( \omega t+\phi \right)$is called the standard wave equation.
Here ‘Y’ represents the displacement of wave particles at time ‘t’. Coefficient of trigonometric function ‘a’ is called the amplitude of the wave.’$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
Now, velocity is $\dfrac{dY}{dt}$, as here, the position is represented by ‘Y’.
So, differentiating the above equations, we get:
${{v}_{1}}=0.1\times 100\pi cos\left( 100\pi t+\dfrac{\pi }{3} \right)$ $\left[ \dfrac{d(sin(ax+b)}{dx}=acos(ax+b) \right]$
${{v}_{2}}=-0.1\times \pi sin\left( \pi t\right)$ $\left[ \dfrac{d(cos(ax+b)}{dx}=-asin(ax+b) \right]$
or ${{v}_{2}}=0.1\times \pi cos\left( \pi t+\dfrac{\pi }{2} \right)$ $\left[ cos(\theta +\dfrac{\pi }{2})=-sin(\theta ) \right]$
Now, ${{v}_{1}}=10\pi cos\left( 100\pi t+\dfrac{\pi }{3} \right)$
And ${{v}_{2}}=0.1 \pi cos\left( \pi t+\dfrac{\pi }{2} \right)$
But, in the question, we are asked about phase difference at t=0:
Hence putting t=0 in velocity equation;
${{v}_{1}}=10\pi cos\left( \dfrac{\pi }{3} \right)$
${{v}_{2}}=0.1 \pi cos\left( \dfrac{\pi }{2} \right)$
Hence we see that phase difference between the two velocities is:
$\dfrac{\pi}{2}-\dfrac{\pi}{3}=\dfrac{\pi}{6}$

So, the correct answer is “Option B”.

Note: Students must not get confused by the actual phase and initial phase ( phase at t=0 ). Actual phase means phase difference between the two particles at any general time i.e. the whole term inside of trigonometric functions. Example: cos($\omega t+\phi$), $\omega t+\phi$ is the actual phase of the particle.