Figure below shows a massless a pulley, a spring of constant k = 1000 N/m and a mass 1 kg. On displacing the mass slightly, if the frequency of its vertical oscillation is $\mathrm{x}\times {10}^1\mathrm{Hz}$ , then find x.

As the mass m moves by a length y , the pulley will shift by y . This should bring an extension of 2y on spring so that a length y can be shared by the spring as well the string. The restoring force on spring is F = 2Ky

Tension on string: ${\mathrm{T}}^1=\mathrm{F} = 2\mathrm{Ky}$

The tension on the string attached to the mass is $\mathrm{T} ={\mathrm{T}}^1=\mathrm{F} = 4\mathrm{Ky}$

$$\begin{gathered} \mathrm{T} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{{\mathrm{k}}_{\mathrm{aq}}}} = 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{4\mathrm{k}}} \\ \mathrm{f} = \frac{1}{\mathrm{T}} = \frac{1}{2\mathrm{\pi }}\sqrt{4000} \approx 10\mathrm{Hz} \approx 1\mathrm{x}{10}^1\mathrm{Hz}\Rightarrow \mathrm{x} = 1 \end{gathered}$$