Q118 / 150 · MHT CET 2016 · Physics

A progressive wave is represented by $y = 12 \sin (5 t - 4 x)$ $cm$ . On this wave, how far away are the two points having a phase difference of $90 °$ ?

A
$\frac{π}{2} c m$
B
$\frac{π}{4} c m$
Correct answer
C
$\frac{π}{8} c m$
D
$\frac{π}{16} c m$

Solution

y = 12 \sin (5 t - 4 x)

Comparing in

y = A \sin (w t - k x)

We have

A = 12, w = 5 a n d k = 4

(w t - k x) =

phase difference

= \frac{π}{2}

∴ | 5 t - 4 x | = \frac{π}{2}

When

t = 0, 4 x = \frac{π}{2}

∴ x = \frac{π}{8} c m

Hence, the correct answer is option: 'C'.

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