Simple Harmonic Motion 5 Question 10
15. For periodic motion of small amplitude $A$, the time period $T$ of this particle is proportional to
(a) $A \sqrt{\frac{m}{\alpha}}$
(b) $\frac{1}{A} \sqrt{\frac{m}{\alpha}}$
(c) $A \sqrt{\frac{\alpha}{m}}$
(d) $\frac{1}{A} \sqrt{\frac{\alpha}{m}}$
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Answer:
Correct Answer: 15. (b)
Solution:
- $[\alpha]=\frac{P E}{x^{4}}=\frac{ML^{2} T^{-2}}{L^{4}}=\left[ML^{-2} T^{-2}\right]$
$$ \begin{aligned} & \therefore \quad \frac{m}{\alpha}=\left[L^{2} T^{2}\right] \\ & \Rightarrow \quad \frac{1}{A} \sqrt{\frac{m}{\alpha}}=[T] \end{aligned} $$
As dimensions of amplitude $A$ is [L].
Hence, the correct option is (b).
