Question
The displacement of an elastic wave is given by the function $\text{y}=3\sin\omega\text{t}+4\cos\omega\text{t}$ where y is in cm and t is in second. Calculate the resultant amplitude.

Answer

$\because\text{y}=3\sin\omega\text{t}+4\cos\omega\text{t}\ ...(\text{i})$Let $3=\text{a}\cos\phi\ ...(\text{ii})$
$4=\text{a}\sin\phi\ ...{\text{iii}}$
Then $\text{y}=\text{a}\cos\phi\sin\omega\text{t}+\text{a}\sin\phi\cos\omega\text{t}$
$\text{y}=\text{a}\sin(\omega\text{t}+\phi) $
From (ii) and(iii)
$\tan \phi=\frac{4}{3}$ or $\phi=\tan^{-1}\frac{4}{3}$
On squaring and adding (ii) and (iii) equations
$\text{a}_2\cos^2\phi+\text{a}^2\sin^2\phi=3^2+4^2$
$\text{a}^2(\cos^2\phi+\sin^2\phi)=9+16$
$\text{a}^2=25\Rightarrow\text{a}=5$
$\text{y}'=5\sin(\omega\text{t}+\phi)$ when $\phi=\tan^{-1}\frac{4}{3}$
Hence, New amplitude is 5 cm.

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