Two waves, travelling in the same direction through the same regi… - Vidyadip

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 4mm and the phase difference between the waves is 90°, what is the resultant amplitude?

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Answer

Phase difference $\phi=\frac{\pi}{2}$ f and $\lambda$ are same. So, $\omega$ is same.$\text{y}_1=\text{r}\sin\text{wt},\ \text{y}_2=\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)$
From the principle of superposition$\text{y}=\text{y}_1+\text{y}_2$
$=\text{r}\sin\text{wt}+\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)$
$=\text{r}\Big[\sin\text{wt}+\text{r}\sin\big(\text{wt}+\frac{\pi}{2}\big)\Big]$
$=\text{r}\Bigg[2\sin\bigg\{\frac{\big(\text{wt}+\text{wt}+\frac{\pi}{2}\big)}{2}\bigg\}\cos\bigg\{\frac{\big(\text{wt}-\text{wt}-\frac{\pi}{2}\big)}{2}\bigg\}\Bigg]$
$\Rightarrow\text{y}=2\text{r}\sin\Big(\text{wt}+\frac{\pi}{4}\Big)\cos\Big(\frac{-\pi}{4}\Big)$
Resultant amplitude $=\sqrt{2}\text{r}=4\sqrt{2}\text{mm}$ (because r = 4mm)

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