Question

A train of sound waves is propagated along an organ pipe and gets reflected from an open end. If the displacement amplitude of the waves (incident and reflected) are 0.002 cm, the frequency is 1000 Hz and wavelength is 40 cm, then the displacement amplitude of vibration at a point at distance 10 cm from the open end inside the pipe, is

• A. 0.002 cm
• B. 0.003 cm
• C. 0.001 cm
• D. 0.000 cm

Answer 1

The correct option is D

0.000 cm

y = 2A cos ($\frac{2\pi x}{\lambda }$) sin ($\omega$ t) , where x = 0 is the open end from where the wave gets reflected.

Amplitude of stationary wave $A_s$ = 2A cos ($\frac{2\pi x}{\lambda }$)

For the position, x = 0.1 m, the amplitude will be, $A_s$ = 2 x 0.002 cos [$\frac{2\pi \left(0.1\right)}{0.4}$] cm = 0 cm