Question

On the superposition of the two waves represented by equation,
$y_1=A\sin (\omega t-kx)$
$y_2=A\cos (\omega t-kx+\frac{\pi }{6})$
The resultant angular frequency of the oscillation will be

• A. $\omega$
• B. $2\omega$
• C. $\frac{\omega }{2}$
• D. $3\omega$

Answer 1

The correct option is A $\omega$

Given: $y_1=A\sin (\omega t-kx)$
$y_2=A\cos (\omega t-kx+\frac{\pi }{6})$
$y_{\text{net}}=y_1+y_2$
$\Rightarrow y_{\text{net}}=A\sin (\omega t-kx)+A\cos (\omega t-kx+\frac{\pi }{6})$
$\Rightarrow y_{\text{net}}=A\sin (\omega t-kx)+A\cos (\omega t-kx)\cos \left(\frac{\pi }{6}\right)-A\sin (\omega t-kx)\sin \left(\frac{\pi }{6}\right)$
$\Rightarrow y_{\text{net}}=A\sin (\omega t-kx)(1-\text{sin}\left(\frac{\pi }{6}\right))+A\cos (\omega t-kx)\text{cos}\left(\frac{\pi }{6}\right)$
$\Rightarrow y_{\text{net}}=\frac{A}{2}\text{sin}(\omega t-kx)+A\cos (\omega t-kx)\text{cos}\left(\frac{\pi }{6}\right)$
$\Rightarrow y_{\text{net}}=A\sin (\omega t-kx)\text{sin}\frac{\pi }{6}+A\cos (\omega t-kx)\text{cos}\frac{\pi }{6}$
$\therefore y_{\text{net}}=A\cos (\omega t-kx-\frac{\pi }{6})$
The angular frequency of resultant oscillation is $\omega$

Note: When two waves having same $\omega$, $k$ and propagating in the same direction are superimposed, the resultant oscillation will have the same angular frequency.