Question

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

Answer 1

Hint: To calculate the phase difference of both particle
First we have to calculate the velocity of a particle by differentiating the displacement of both particles with respect to time. And then compare the equations by the standard equation of simple harmonic motion .
\[x=A\sin \left( wt+\phi \right)\]

Complete step-by-step answer:
As given in the question, the motion of both particle given as:
\[{{y}_{1}}=0.1\sin \left( 100\pi +\dfrac{\pi }{3} \right)\text{ }...........\text{ 1}\]
\[{{y}_{2}}=0.1\cos \pi t\text{ }........\text{ 2}\]
For finding the velocity of each particle, we have to differentiate the equation of motion with respect to time.
As we know, the rate of change of displacement is velocity.
So,
\[{{v}_{1}}=\dfrac{d{{y}_{1}}}{dt}=\dfrac{d}{dt}\left[ 0.1\sin \left( 100\pi t+\dfrac{\pi }{3} \right) \right]\]
\[\Rightarrow {{v}_{1}}=0.1\times 100\pi \cos \left( 100\pi +\dfrac{\pi }{3} \right)\]
\[\Rightarrow {{v}_{1}}=10\pi \cos \left( 100\pi t+\dfrac{\pi }{3} \right)\text{ }.........\text{ 3}\]
We know as given
\[{{y}_{2}}=0.1\cos \pi t\]
\[\Rightarrow {{y}_{2}}=0.1\sin \left( \pi t+\dfrac{\pi }{2} \right)\]
\[{{v}_{2}}=\dfrac{d{{y}_{2}}}{dt}=0.1\pi \cos \left( \pi t+\dfrac{\pi }{2} \right)\]
\[\Rightarrow {{v}_{2}}=0.1\pi \cos \left( \pi t+\dfrac{\pi }{2} \right)\text{ }.........\text{ 4}\]
Equation of particle completing simple harmonic
Motion is
\[x=A\sin \left( wt+\phi \right)\]
\[v=\dfrac{dx}{dt}=Aw\cos \left( wt+\phi \right)\]
Where \[\phi =\text{Phase of motion}\]
At\[t=0\], phase of motion \[=\phi \]
By equation 3 at \[t=0\] phase of velocity of particle ‘1’ is \[{{\phi }_{1}}=\dfrac{\pi }{3}\]
By equation 4 at \[t=0\] phase of velocity of particle ‘2’ is \[{{\phi }_{2}}=\dfrac{\pi }{2}\]
Thus, phase difference of the velocity of particle ‘1’ with respect to the velocity of particle ‘2’ is at \[t=0\Rightarrow \text{Phase of 1st particle at t}=0-\text{Phase of 2nd particle at t}=0\]
\[=\dfrac{\pi }{3}-\dfrac{\pi }{2}=\dfrac{-\pi }{6}\]
Hence, the phase difference of the velocity of particle ‘1’ with respect to the velocity of particle ‘2’ at \[t=0\] is \[\dfrac{-\pi }{6}\]
Therefore, the correct choice is: (C) \[\dfrac{-\pi }{6}\]

Note: Phase difference is used to describe the difference in between two waves.
For simple harmonic wave first need to calculate the velocity of each particle then compose it with standard equation of SHM which is
\[x=A\sin \left( wt+\phi \right)\]
\[v=Aw\cos \left( wt+\phi \right)\]
\[a=-A{{w}^{2}}\sin \left( wt+\phi \right)\]