1.

A mass m is suspended to a spring of length L and force constant k. The frequency of vibration is f_(1). The spring is cut into two equal parts and each half is loaded with same mass m. The new frequency f_(2) is given by :

Answer»

`f_(2)=sqrt(2)f_(1)`
`f_(2)=(f_(1))/(sqrt(2))`
`f_(2)=f_(1)//2`
`f_(2)=(sqrt(2))/(f_(1))`.

Solution :Here `f_(1)=(1)/(2pi)sqrt((k_(1))/(m))`
and `""k_(1)=(mg)/(DeltaL)`.
Now `""k_(2)=(mg)/(DeltaL//2)=2k_(1)`
`f_(2)=(1)/(2pi)sqrt((2k_(1))/(m))`
and `f_(2)=sqrt(2)f_(1)`.
Correct CHOICE is (a).


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