Simple Harmonic Motion Ques 50
- 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) $\frac{1}{A} \sqrt{\frac{\alpha}{m}}$
Show Answer
Answer:
Correct Answer: 50.(b)
Solution:
Formula:
- $[\alpha]=[\frac{P E}{x^{4}}]=[\frac{ML^{2} T^{-2}}{L^{4}}]=\left[ML^{-2} T^{-2}\right]$
$ \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).
BETA
