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Wave Motion and Waves on a String — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Motion and Waves on a String3 Marks
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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