Question

Two SHM’s are respectively by $y = {a_1}\sin \left( {\omega t - kx} \right)$ and $y = {a_2}\cos \left( {\omega t - kx} \right)$. The phase difference between the two is
A. $\dfrac{\pi }{2}$
B. $\dfrac{\pi }{4}$
C. $\dfrac{\pi }{6}$
D. $\dfrac{{3\pi }}{4}$

Answer 1

Use the formula for the phase difference of the two waves. This formula gives the relation between the phase difference between two SHM’s, phase of the first SHM and phase of the second SHM. Compare these two equations of SHM with the standard equation of SHM and determine the phase of the two SHM’s. Then substitute the values of phase of two SHM’s in the formula to calculate the required phase difference between the two SHM’s.

Complete step by step solution:
We have given that the two simple harmonic motions are represented by the equations
$y = {a_1}\sin \left( {\omega t - kx} \right)$ …… (1)
$y = {a_2}\cos \left( {\omega t - kx} \right)$ …… (2)
We have asked to calculate the phase difference between these two simple harmonic motions.We know that the equation for motion of a particle in simple harmonic motion is given by
$y = a\sin \left( \phi \right)$ …… (3)
Here, $a$ is the amplitude of the motion and $\phi$ is the phase of the particle's simple harmonic motion.
We can write the equation (2) as
$y = {a_2}\sin \left( {\omega t - kx + \dfrac{\pi }{2}} \right)$ …… (4)
Compare equation (1) with the equation (3), we get
${\phi _1} = \omega t - kx$
Compare equation (4) with the equation (3), we get
${\phi _2} = \omega t - kx + \dfrac{\pi }{2}$
The phase difference $\Delta \phi$ between the two simple harmonic motions of the particle is given by
$\Delta \phi = {\phi _2} - {\phi _1}$
Here, ${\phi _2}$ is the phase difference of the second SHM and ${\phi _1}$ is the phase difference of the first SHM.
Substitute $\omega t - kx + \dfrac{\pi }{2}$ for ${\phi _2}$ and $\omega t - kx$ for ${\phi _1}$ in the above equation.
$\Delta \phi = \omega t - kx + \dfrac{\pi }{2} - \left( {\omega t - kx} \right)$
$\Rightarrow \Delta \phi = \omega t - kx + \dfrac{\pi }{2} - \omega t + kx$
$\therefore \Delta \phi = \dfrac{\pi }{2}$
Therefore, the phase difference between the two equations representing simple harmonic motion is $\dfrac{\pi }{2}$.

Hence, the correct option is A.

Note: The students should not forget to convert the equation of motion of the second particle in the simple harmonic motion from cosine of the phase to sine of the phase. If this conversion is not done then one cannot compare this equation of motion with the standard equation of motion of the particle in simple harmonic motion.