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Two equal masses are connected by a spri...

Two equal masses are connected by a spring satisfying Hook's law and are placed on a frictionless table. The spring is elongated a little and allowed to go. Let the angular frequency of oscillations be `omega`. Now one of the masses is stopped. The square of the new angular frequency is `:`

A

`omega^(2)`

B

`(omega^(2))/(2)`

C

`(omega^(2))/(3)`

D

`2omega^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B

`sqrt((k)/(mu))=omega mu=(m_(1)m_(2))/(m_(1)+m_(2))`
`omega=sqrt((2k)/(m))`
`omega_(2)=sqrt((k)/(m))`
`omega_(2)^(2)=(k)/(m)=(omega^(2))/(2) Ans (B)`
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