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Q.
A force of $6.4\, N$ stretches a vertical spring by $0.1 \,m$. The mass that must be suspended from the spring so that it oscillates with a time period of $\pi / 4$ second.
Oscillations
Solution:
Spring constant $K=\frac{6.4}{0.1}=64\, N / m$.
Now $T=2 \pi \sqrt{\frac{m}{k}}$
or $\frac{\pi}{4}=2 \pi \sqrt{\frac{m}{64}} \therefore m=1\, kg$