When two progressive waves $y_{1}=4 \sin (2 x-6 t)$ and $y_{2}=3 \sin \left(2 x-6 t-\frac{\pi}{2}\right)$ are superimposed, the amplitude of the resultant wave is

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Waves & Sound

When two progressive waves

y_{1} = 4 sin (2 x - 6 t)

and

y_{2} = 3 sin (2 x - 6 t - \frac{π}{2})

are superimposed, the amplitude of the resultant wave is

Answer: 5

Solution

The two waves have the same frequency and wave number, differing only in amplitude and phase.

y_{1} = 4 sin (2 x - 6 t)

y_{2} = 3 sin (2 x - 6 t - π /2) = - 3 cos (2 x - 6 t)

Resultant:

y = y_{1} + y_{2} = 4 sin (2 x - 6 t) - 3 cos (2 x - 6 t)

This can be written as

y = A sin (2 x - 6 t + ϕ)

where:

A = 4^{2} + 3^{2} = 16 + 9 = 25 = 5

The amplitude of the resultant wave is 5.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer

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