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The length of a shown in figure betweenn...

The length of a shown in figure betweenn the pulleys is `1.5 m` and its mass is `15 g`. Find the freqency of vibration with which the wire vibrates in four loops leaving the middle point of the wire between the pulleys at rest. `(g = 10 m//s^(2))`

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The correct Answer is:
`(400)/(3) Hz`
`(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`.
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