Q130 / 160 · TS EAMCET 2023 (13 May Shift 2) · Physics
A force of $6.4 \mathrm{~N}$ stretches a vertical spring by $0.1 \mathrm{~m}$. If it were to oscillate with a period of $\frac{\pi}{4}$ then the mass that is to be suspended from the spring is
- A$\frac{\pi}{4} \mathrm{~kg}$
- Correct answerB$1 \mathrm{~kg}$
- C$\frac{1}{\pi} \mathrm{kg}$
- D$10 \mathrm{~kg}$
Solution
Time period, $\mathrm{T}=\frac{\pi}{4}$
Force $\mathrm{F}=6.4 \mathrm{~N}$ $\mathrm{x}=0.1 \mathrm{~m}$
Spring force, $\mathrm{F}=\mathrm{kx}$
$$
\begin{aligned}
& 6.4=k \times 0.1 \\
& k=64
\end{aligned}
$$
Time period, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$
$$
\frac{\pi}{4}=2 \pi \sqrt{\frac{\mathrm{m}}{64}}
$$
mass, $\mathrm{m}=1 \mathrm{~kg}$.
Hence, the correct answer is option: 'B'.