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)

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