Question

A string $120cm$ in length sustains a standing wave, with the points of string at which the displacement amplitude is equal to $2mm$ being separated by $15.0cm$. The maximum displacement amplitude is

• A. $4mm$
• B. $2.2mm$
• C. $4.2mm$
• D. $2mm$

Answer 1

Answer: (D)

$2mm$

From figure, points A,B,C,D,E and F have equal displacement amplitudes.

$x_E-x_A=\lambda =4\times 15cm=60cm$

Thus $n=\frac{120}{60/2}=4$

Thus it corresponds to the 4th harmonic.

Distance of node from point $A=7.5cm$.
Amplitude of stationary wave can be written as $a=A\sin Kx$

Here, $a=\sqrt{2}mm$, $K=\frac{2\pi }{\lambda }=\frac{2\pi }{60}$ and $x=7.5cm$

$⟹\sqrt{2}=A\sin (\frac{2\pi }{60}\times 7.5)=Asin\frac{\pi }{4}=\frac{A}{\sqrt{2}}$

$⟹A=2mm$