A train of sound wsves is propagated along a wide pipe and it is reflected from an open end. If the amplitude of the waves is 0.002cm, the frequency 1000Hz and the wvelength 40cm, the amplitude of viberation at a point 10cm from open end inside the pipe will be(0.0.x)cm. Find teh value of x.

Options

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

Answer

D

Explanation

The equation of stationary wave foe open organ pipe can be written as
` y = 2 A cos (( 2 pi x)/(lambda)) sin (( 2 pi f t)/(v))`
where `x = 0` is the open end from where the wave gets reflected.
Amplitude of stationary wave is
`A_(s) = 2 A cos (( 2pi x)/(lambda))`
For ` x = 0.1 m`,
`A_(s) = 2 xx 0.02 cos [( 2pi xx 0.1)/(0.4)] = 0`