Question

Two waves, travelling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude?

Answer 1

Given,
Phase difference between the two waves travelling in the same direction,
\[\phi = 90^\circ\ = \frac{\pi}{2}\]
Frequency f and wavelength
\[\lambda\] are the same. Therefore, ω will be the same.
Let the wave equations of two waves be: 
\[y_1  = r\sin\omega t\] 
\[y_2  = r\sin\left( \omega t + \frac{\pi}{2} \right)\]
Here, r is the amplitude.
From the principle of superposition, we get:

\[y =  y_1  +  y_2 \] 

\[     = r\sin\omega t + r\sin\left( \omega t + \frac{\pi}{2} \right)\] 

\[     = r  \left[ \sin\omega t + \sin\left( \omega t + \frac{\pi}{2} \right) \right]\] 

\[     = r\left[ 2\sin\left\{ \left( \frac{\omega t + \omega t + \frac{\pi}{2}}{2} \right) \right\}\cos\left\{ \left( \frac{\omega t - \omega t - \frac{\pi}{2}}{2} \right) \right\} \right]\] 

\[     = 2r\sin\left( \omega t + \frac{\pi}{4} \right)\cos\left( - \frac{\pi}{4} \right)\] 

\[     = \sqrt{2}r\sin\left( \omega t + \left( \frac{\pi}{4} \right) \right)\]
∴ Resultant amplitude,

\[r' = \sqrt{2}r = 4\sqrt{2}  \text{ mm }\]