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

Two open pipes have lengths l and l + trianglel. Beat frequency in the

1.	Two open pipes have lengths l and l + trianglel. Beat frequency in their first mode would be,
Answer» `(V)/(2l)` `(v)/(4l)` `(v)/(2l ^(2)) (Delta l) ^(2)` `(v )/(2l ^(2)) Delta l` Solution :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)`