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 for both of them, the frequency is 100 Hz and wavelength is 40 cm, then the displacement amplitude of vibration at a point at a distance 10 cm from the open end 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

The equation of stationary wave for open organ pipe can be written as
$y=2Acos\left(\frac{2\pi }{\lambda }x\right)sin\left(\frac{2\pi ft}{\lambda }\right)$, here x = 0 is the open end from where the wave gets reflected.
Amplitude of stationary wave is $A_s=2Acos\left(\frac{2\pi x}{\lambda }\right)$
For x = 0.1 m, $A_s=2\times 0.002cos\left[\frac{2\pi \times 0.1}{0.4}\right]=0$