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A mass of 0.2 hg is attached to the lowe...

A mass of 0.2 hg is attached to the lower end of a massles spring of force constant `200(N)/(m)` the opper end of which is fixed to a rigid support. Study the following statements.

A

In equilibrium , the spring will be stretched by 1 cm

B

If the mass is raised till the spring becomes unstretched and then released , it will go down by 2 cm before moving upwards

C

The frequency of oscillation will be nearly 5 Hz

D

All of the above

Text Solution

Verified by Experts

The correct Answer is:
D
(d) At equilibrium , `x=(mg)/(k)=(0.2xx10)/(200)=0.01 m = 1cm `
`f=(1)/(2pi)sqrt((K)/(m)) =(1)/(2pi)sqrt((200)/(0.2))~~ 5 Hz`
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