A knife edge divides a sonometer wire in two parts which differ in length by 2mm. The length of the wire is 1m. The two parts of the string when sounded together produce on beat per second, then the frequency of the smaller and the longer parts of the wire in Hz will be

Options

A. 250.5 and 249.5
B. 249.5 and 250.5
C. 124.5 and 125.4
D. 125.4 and 124.5

Answer

A

Explanation

`l_1-l_2 = 2xx10^(-3)`
`l_1+l_2 = 1`
`therefore 2l_1=1.002 therefore l_1 = =0.501m1`
`l_2=0.499m`
`n_2-n_1 = 1 " " ….(i)`
`n_1l_1=n_2l_2`
`n_1/n_2=l_2/l_2=499/501 therefore n_1=499/509n_2`
The eqn (i) becomes,
`n_2-499/501 n_2 = 1 therefore 2n_2 = 501`
`n_2 = 205.5 Hz " and " n_1 =205.5-1=249.5`