Question

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. $\frac{1}{A}\sqrt{\frac{\alpha }{m}}$

Answer 1

Answer: (B)

$[\alpha ]=\left[\frac{PE}{x^4}\right]=\left[\frac{M{L}^2{T}^{-2}}{L^4}\right]=[M{L}^{-2}{T}^{-2}]$
$\therefore \left[\frac{m}{\alpha }\right]=[L^2{T}^2]\Rightarrow \therefore \left[\frac{1}{A}\sqrt{\frac{m}{\alpha }}\right]=[T]$
As dimensions of amplitude A is [L].
Hence, the correct option is (b).