Question

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

Answer 1

Given,

The displacement of an elastic wave is given by the function

$y=3\sin \omega t+4\cos \omega t=y_1+y_2$

Where $y_1=a_1\sin \omega t=3\sin \omega t{y}_2=a_2\cos \omega t=4\cos \omega t$

$=4\sin (\omega t+\frac{\pi }{2})$

where $y$ is in $cm$

Phase difference, $\Delta \varphi =\frac{\pi }{2}$

Resultant amplitude of given wave

$a=\sqrt{a_1^2+a_2^2+2a_1{a}_2\cos \Delta \varphi }$

$=\sqrt{a_1^2+a_2^2}$

$a=\sqrt{3^2+4^2}=5cm$

Final Answer: $5cm$