A loudspeaker that produces signals from `50` to `500 Hz` is placed at the open end of a closed tube of length `1.1 m`. The lowest and the highest frequency that excites resonance in the tube are `f_(1) and f_(h)` respectively . The velocity of sound is `330 m//s`. Then

Options

A. `f_(1) = 50 Hz`
B. `f_(h) = 500 Hz`
C. `f_(1) = 75 Hz`
D. `f_(h) = 450 Hz`

Answer

C, D

Explanation

For a closed tube
`f_(n) = ( nv)/( 4 L)`
`L = 1.1 m , v = 330 m//s`
`f_(n) = ( n xx 330)/( 4 xx 1.1) = 500 Hz rArr n = 6.66`
highest frequency ,
`f_(h) = ( 6 xx 330)/( 4 xx 1.1) = 450 Hz`
Lowest frequency ,
`f_(1) = ( 1 xx 330)/(4.4) = 75 Hz`