The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 15 g. what is the frequency of vibration with which the wire vibrates in four loops leaving the middle point of the wire between the pullys at rest ? `(g=10 ms^(2))`

Answer

`(400)/(3) Hz`

Explanation

`(4lambda)/(2) = 1.5`
`lambda = 0.75 m`
`v = sqrt((10 xx 10 xx 1.5)/(15 xx 10^(-3))) = 100 m//s`
`f = (v)/(lambda) = (100)/(0.75) = (400)/(3) Hz`.