Question

A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.

Find the frequency of oscillation?

Answer 1

The time period of the oscillating system depends on the mass-spring constant given by T = `2pisqrt(m/k)`

It does not depend on the amplitude.

But from equation (iii),

`(2mg)/k = x`

`(2mg)/k = 4  cm = 4 xx 10^-2  m`

∴ `m/k = (4 xx 10^-2)/(2g) = (2 xx 10^-2)/g`

∴ `k/m = g/(2 xx 10^-2)`

And v = frequency = `1/T = 1/(2pi) sqrt(k/m)`

∴ v = `1/(2 xx 3.14) sqrt(g/(2 xx 10^-2))`

= `1/(2 xx 3.14) sqrt(4.9/10^-2)`

= `1/6.28 xx sqrt(4.9 xx 100)`

= `10/6.28 xx 2.21`

= 3.51 Hz