Question

Find the spring constant of the spring if a spring of length `2m` and mass is `82` grams, is stretched to a length of `2.8 m`. Its frequency is now `25 Hz` after it has been stretched?

Answer 1

The spring constant is `=2023.3Nm^-1`

Explanation:

The frequency is `f=25Hz`

The period is

`T=1/f=1/25=0.04s`

The relationship is

`T=2pisqrt(m/k)`

The mass is `m=0.082kg`

Therefore,

`T^2=4pi^2m/k`

`k=(4pi^2m)/(T^2)`

`k=(4*pi^2*0.082)/(0.04^2)=2023.3Nm^-1`

Answer 2

Restoring force acting due to compression of a spring is given as, `F= -Kx` (where, `K` is the spring constant),which can be compared with equation of force of S.H.M `F= - m (omega)^2 x`

So, comparing we get, `K= m(omega)^2 = m(2 pi n)^2` (where, `n` = frequency)

Given, `m = 82/1000 Kg` and. `n=25 Hz`

putting the values,we get `K=2023.3N/m`