A tunning fork produces 4 beats/s both with 50 and 40 cm of a stretched wire of sonometer. The frequency of fork is :

Options

A. 36 Hz
B. 50 Hz
C. 90 Hz
D. 110 Hz

Answer

A

Explanation

For stretched string `V_(1) - (1)/(2 l_(1)) sqrt((T)/(m))`
Then `v. = ((V)/(V - U_(s)) )` V
`rArr 1000 = ((350)/(350 - 50)) v rArr v = (1000 xx 300 )/(350 )`
When the source is moving away from the observer then
` v.. = ((V)/(V + U_(s)) ) ` V
` rArr " " v.. = ((350)/(350 + 50) ) xx (1000 xx 300)/(350)`
v.. = 750 Hz
correct choice is (d).
` v_(2) = (1)/(2l_(2) sqrt((T)/(m))`
`therefore (v_(1))/(v_(2)) = (l_(2))/(l_(1)) = (40)/(50) = (4)/(5) " " rArr " " V_(1) = (4)/(5) v_(2)`
Let the unknonw frequency of tunning fork = v
`therefore " " v- v_(1) = 4 `
`v_(2) - v = 4 `
Adding `" " v_(2) - v_(1) = 8 `
`v_(2) - (4)/(5) v_(2)= 8 `
`(1)/(5) v_(2) = 8 " " rArr v_(2) = 40` Hz.
`therefore " " 40 - v = 4 " " rArr " " v = 36 ` Hz.
Hence the correct choice is (a) .