Two SHMs are respectively represented by $y_{1} = a\ sin(\omega t - kx)$ and $y_{2} = b\cos{(\omega t - kx})$ The phase difference between the two is

Question

Dr. Abubacker · Accepted Answer

(c) $y_{1} = a\sin{(\omega t - kx)}$;
$y_{2} = b\cos\left( \omega t - \frac{k}{x} ight) = b\sin\left( \omega t - \frac{k}{x} + \pi/2 ight)$
$	herefore$ Phase difference$= \left( \omega t - \frac{k}{x} + \pi/2 ight) - (\omega t - kx)$
$= \pi/2$

Q. Two SHMs are respectively represented by $y_{1} = a\ sin(\omega t - kx)$ and $y_{2} = b\cos{(\omega t - kx})$ The phase difference between the two is

Explanation

Reviewed by Dr. Abubacker

Want more practice?