Question

Two particles are executing simple harmonic motion. At an instant of time $t$ their displacements are
and $y_1=a\cos (\omega t)$
$y_2=a\sin (\omega t)$
Then the phase difference between $y_1$ and $y_2$ is:

• A. $120^{\circ }$
• B. $90^{\circ }$
• C. $180^{\circ }$
• D. zero

Answer 1

Answer: (B)

Cosine expression can be expressed in terms of sine expression using identity
$\sin (\pi /2+\theta )=\cos \theta$.
The given displacement equations are
$y_1=a\cos \omega t$
$y_1=a\sin \left(\omega t+\frac{\pi }{2}\right)$
and $y_2=a\sin \omega t$
Hence, phase difference between $y_1$ and $y_2$
$=\left(\omega t^{\prime }+\frac{\pi }{2}\right)-\omega t=\frac{\pi }{2}$
Hence, phase difference is $90^{\circ }.$