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Waves — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsWaves5 Marks
Question
Prove that if a, and a, are the amplitudes of two interfering waves $\phi$ and is their phase difference, the amplitude of the resultant wave is given by,
$\text{a}^2=\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi$
✓

Answer

Let $\text{y}_1=\text{a}_1\cos\ \omega\text{t},$
$\text{y}_2=\text{a}_52\cos\ (\omega\text{t}+\phi)$
be the two displacements.
On superposition,
$\text{y}=\text{y}_1+\text{y}_2$
$=\text{a}_1\cos\ \omega\text{t}+\text{a}_2\cos(\omega\text{t}+\phi)$
$=\text{a}_1\cos\omega\text{t}+\text{a}_2\cos\ \omega\text{t}\cos\phi$
$=-\text{a}_2\sin\omega\text{t}\sin\phi$
$=-(\text{a}_2\sin \phi)\sin\omega\text{t}\dots(1)$
$\text{put}\text{ a}_1+\text{a}_2\cos\phi=\text{a}\cos\theta\dots(2)$
$\text{a}_2\sin\phi=\text{a}\sin\theta\dots(3)$
Squaring and adding (2) and (3), we have
$\text{y}=\text{a}\cos\theta\cos\omega\text{t}-\text{a}\sin\theta\sin\omega\text{t}$
$=\text{a}\cos(\omega\text{t}+\theta)$
$\therefore\text{y}=\Big(\text{a}^2_1+\text{a}^2_2+2\text{a}_1\text{a}_2\cos\phi\Big)^{\frac{1}{2}}\cos(\omega\text{t}+\theta)$
where $\theta=\tan^{-1}\Big(\frac{\text{a}_2\sin\phi}{\text{a}_1+\text{a}_2\cos\phi}\Big)$

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