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 $y_{2}$
$=\left(\omega t'+\frac{\pi}{2}\right)-\omega t=\frac{\pi}{2}$
Hence, phase difference is $90^{\circ} .$