Question

An oscillator is formed of a mass m = 2 kg attached to a spring having a spring constant k = 200 N/m. The frequency of oscillation is equal to:

Answer 1

Given,

The mass attached to the spring, m=2 kg

The spring constant of the spring to which the mass is attached, k=200 N/m

The time of the oscillation of the oscillator is given by,

`T=2\pi\sqrt[]{(m)/(k)}`

The frequency is related to the period of the oscillator as,

`f=(1)/(T)`

Thus the frequency of the oscillator is given by,

`f=(1)/(2\pi)\sqrt[]{(k)/(m)}`

On substituting the known values in the above equation,

`\begin{gathered} f=(1)/(2\pi)\sqrt[]{(200)/(2)} \\ =1.59\text{ Hz} \end{gathered}`

Thus the frequency of oscillation of thus formed oscillator is 1.59 Hz