Question

A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?

Answer 1

It is given that:
Energy stored in the spring, E = 5 J
Frequency of the mass-spring system, f = 5
Extension in the length of the spring, x = 25 cm = 0.25 m
$$\begin{gathered} \mathrm{Time}\mathrm{period},T=\frac{1}{5}\mathrm{s} \\ \mathrm{Potential}\mathrm{energy}\left(\mathrm{U}\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ U=\frac{1}{2}k{x}^2 \\ \Rightarrow \frac{1}{2}k{x}^2=5 \\ \Rightarrow \frac{1}{2}k{\left(0.25\right)}^2=5 \\ \Rightarrow k=160\mathrm{N}/\mathrm{m} \\ \mathrm{Time}\mathrm{period}\mathrm{of}\mathrm{spring}\mathrm{mass}\mathrm{system}\mathrm{is}\mathrm{given}\mathrm{by}, \\ T=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)} \\ \mathrm{where}m\mathrm{is}\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{body}\mathrm{hange}d,\mathrm{and} \\ k\mathrm{is}\mathrm{the}\mathrm{spring}\mathrm{constant}. \\ \mathrm{On}\mathrm{substituting}\mathrm{the}\mathrm{respective}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{expression},\mathrm{we}\mathrm{get}: \\ \frac{1}{5}=2\mathrm{\pi }\sqrt{\left(\frac{m}{160}\right)} \\ \Rightarrow m=0.16\mathrm{kg} \end{gathered}$$