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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))`

A

`100/3 Hz`

B

`200/3 Hz`

C

`400/3 Hz`

D

`500/3 Hz`

Text Solution

Verified by Experts

The correct Answer is:
C
`(4lambda)/2=1.5`
`lambda=0.75 cm`
`v=sqrt((10xx10xx1.5)/(15xx10^(-3)))=100 m//s`
`f=v/(lambda)=100/0.75 =400/3 Hz`
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