Two $S H M^{'} s$ are respectively represented by $y = a sin (w t - k x)$ and $y = b cos (w t - k x) .$ Find out the phase difference between the two $?$

90^{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $90^{0}$
$Y = a s i n (w t - k x)$ and
$y = b c o s (w t - k x)$
or $b s i n (w t - k x + 90^{0})$
so $,$ the phase difference is $90^{0}$
Hence,
option $(A)$ is correct answer.

Suggest Corrections

Similar questions

y = a sin (w t - k x)

y = b cos (w t - k x)

S H M^{'} s

y = a sin (ω t - k x)

y = b cos (ω t - k x)

Two waves represented by $y$ = $a s i n (w t - k x)$ and $y$ = $a c o s (w t - k x)$ are superposed. The resultant wave will have an amplitude

y_{1} = A_{1} sin (w t - k x)

y_{2} = A_{2} cos (w t - k x + π / 3)

Join BYJU'S Learning Program