SATHEE: Simple Harmonic Motion 5 Question 10

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) $A \sqrt{\frac{m}{α}}$

(b) $\frac{1}{A} \sqrt{\frac{m}{α}}$

(d) $\frac{1}{A} \sqrt{\frac{α}{m}}$

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Answer:

Correct Answer: 15. (b)

Solution:

$[α] = \frac{P E}{x^{4}} = \frac{M L^{2} T^{- 2}}{L^{4}} = [M L^{- 2} T^{- 2}]$

$\begin{aligned} ∴ \frac{m}{α} = [L^{2} T^{2}] \\ \Rightarrow \frac{1}{A} \sqrt{\frac{m}{α}} = [T] \end{aligned}$

As dimensions of amplitude $A$ is [L].

Hence, the correct option is (b).

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Simple Harmonic Motion 5 Question 10

15. For periodic motion of small amplitude A, the time period T of this particle is proportional to

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15. For periodic motion of small amplitude $A$ , the time period $T$ of this particle is proportional to