Q. Two simple pendulums first of bob mass $M_{1}$ and length $L_{1}$ second of bob mass $M_{2}$ and length $L_{2}.$ $M_{1} = M_{2}$ and $L_{1} = 2L_{2}.$ If the vibrational energy of both is same. Then which is correct
A) Amplitude of $B$ greater than $A$
B) Amplitude of $B$ smaller than $A$
C) Amplitude will be same
D) None of these
Explanation
(b) $n = \frac{1}{2\pi}\sqrt{\frac{g}{l}} \Rightarrow n \propto \frac{1}{\sqrt{l}} \Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{L_{2}}{2L_{2}}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{\sqrt{2}} \Rightarrow n_{2} = \sqrt{2}n_{1} \Rightarrow n_{2} > n_{1}$
Energy$\ E = \frac{1}{2}m\omega^{2}a^{2} = 2\pi^{2}mn^{2}a^{2}$
$\Rightarrow \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{m_{2}n_{2}^{2}}{m_{1}n_{1}^{2}}\ \ \ \ \ \ \ \ \ \lbrack\because E\ is\ same\rbrack$
Given $n_{2} > n_{1}$ and $m_{1} = m_{2} \Rightarrow a_{1} > a_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.