Question

A long and light helical spring is stretched 4.68cm when a given mass is hung on it. what is the period of vibration of the mass if pulled down a little and then released?

Answer 1

`0.44 s`

Explanation:

Suppose, when you stretch the spring by a distance of `x`,restoring force `F` acting is `F=-Kx` (where, `K` is the spring constant)

So,we can compare this with the equation of S.H.M, i,e `F=-m omega^2 x`

So, `omega = sqrt(K/m)` (where, `m` is the mass of the substance hanged,as here,stretching is occurring due to the mass hung from it)

So,period is `T= (2pi)/(omega) = 2pi sqrt(m/K)`

Now,given, when a mass is hung on it,it gets stretched by `4.68 cm`,and here the force acting is the weight of the mass,which is stretching it.

So,we can say, `mg = K(4.68/100)`

or, `m/K = 4.68/100 *(1/g)`

or, `sqrt(m/K) = 0.07`

So, `T= 2pi *0.07 =0.44 s`