Two open pipes have lengths `l and l + trianglel.` Beat frequency in their first mode would be,

Options

A. `(v)/(2l)`
B. `(v)/(4l)`
C. `(v)/(2l ^(2)) (Delta l) ^(2)`
D. `(v )/(2l ^(2)) Delta l`

Answer

D

Explanation

Beat frequency, `f _(1) - f _(2) =(v)/(2l) - (v)/( 2 (l + Delta l ))`
`therefore f _(1) - f _(2) = (v)/(2) [ (1)/(l) -(1)/( l+ Delta l )]`
`=(v)/(2) [ (1 + Delta l -l)/(l ( l + Delta l ))]`
`= (v)/(2) [(Delta l )/(l ^(2) + l Delta l )]`
When `Delta l` is extremely small, `l Delta l lt lt lt lt l^(2)` ans so it can be neglected from the denominator on R.H.S.
Hence, beat frequency is, `f _(1) -f _(2) = (v)/(2l ^(2)) (Delta l)`