Question

A standing wave pattern of amplitude $A$ in a string of length $L$ shows $2$ nodes (plus those at two ends). If one end of the string corresponds to the origin and $v$ is the speed of progressive wave, the disturbance in the string, could be represented (with appropriate phase) as :

• A. $y(x,t)=A\sin \left(\frac{2\pi x}{L}\right)\cos \left(\frac{2\pi vt}{L}\right)$
• B. $y(x,t)=A\cos \left(\frac{3\pi x}{L}\right)\sin \left(\frac{3\pi vt}{L}\right)$
• C. $y(x,t)=A\cos \left(\frac{4\pi x}{L}\right)\cos \left(\frac{4\pi vt}{L}\right)$
• D. $y(x,t)=A\sin \left(\frac{3\pi x}{L}\right)\cos \left(\frac{3\pi vt}{L}\right)$

Answer 1

The correct option is D $y(x,t)=A\sin \left(\frac{3\pi x}{L}\right)\cos \left(\frac{3\pi vt}{L}\right)$

From the given information, there are four nodes of the wave($n=4$), including the two at the ends. Clearly, it forms a sine wave. With amplitude A, wave number, $k=(n-1)\pi /L=3\pi /L.$Writing this in the standard sinusoidal standing wave form, $y=A\sin \left(kx\right)\cos \left(\omega t\right)$,
$\omega =kv$
$y=A\sin \left(3\pi x/L\right)\cos \left(3\pi vt/L\right)$