Question

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 2cm before moving upwards.
C. The frequency of oscillation will be nearly 5 Hz.
D. If the system is taken to the moon, the frequency of oscillation will be the same as that on the earth.

Answer 1

Correct Answer - A::B::C::D
If in equilibrium position elongation of the spring is equal to `x_0` then
`Kx_0=mg` or `x_0=(mg)/(K)=1cm`
If the block is raised till spring becomes unstretched an dthen relesed, then during subsequent motion maximum elongation of the spring (y) will be calculated by energy conservation law.
Loss of `PE` of block `(mgy)=`strain energy `((1)/(2)Ky^2)`
`y=(2mg)/(K)=2cm`
Hence option (b) is correct.
Frequency of oscillation will be.
`f=(1)/(2pi)sqrt((K)/(m))=5Hz`
Hence option (c ) is correct.
Since frequency, `f`, does not depend upon gravitational acceleration therefore frequency will remain unchanged, even if the system is taken to moon. Hence option (d) is also correct.