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, $\varphi =90°=\frac{\pi }{2}$.
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:
$$\begin{gathered} y=y_1+y_2 \\ =r\sin \omega t+r\sin \left(\omega t+\frac{\pi }{2}\right) \\ =r\left[\sin \omega t+\sin \left(\omega t+\frac{\pi }{2}\right)\right] \\ =r\left[2\sin \left\{\left(\frac{\omega t+\omega t+{\displaystyle \frac{\pi }{2}}}{2}\right)\right\}\cos \left\{\left(\frac{\omega t-\omega t-{\displaystyle \frac{\pi }{2}}}{2}\right)\right\}\right] \\ =2r\sin \left(\omega t+\frac{\pi }{4}\right)\cos \left(-\frac{\pi }{4}\right) \\ =\sqrt{2}r\sin \left(\omega t+\left(\frac{\pi }{4}\right)\right) \end{gathered}$$
∴ Resultant amplitude, $r\text{'}=\sqrt{2}r=4\sqrt{2}\mathrm{mm}$