Question

Two simple pendulums first of bob mass $M_1$ and length $L_1$, second of bob mass $M_2$ and length $L_2.M_2=M_{2}and{L}_1=2L_2.$ If the vibrational energies of both are 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 ofthe above

Answer 1

Answer: (B)

Frequncey, $n=\frac{1}{2\pi }\sqrt{\frac{g}{l}}$
$orn\propto =\frac{1}{\sqrt{l}}$
$\therefore \frac{n_1}{n_2}=\sqrt{\frac{l_2}{l_1}}=\sqrt{\frac{L_2}{2L_2}}$
$\frac{n_1}{n_2}=\frac{1}{\sqrt{2}}$
$n_2=\sqrt{2}{n}_1$
$\Rightarrow n_2>n_1$
Energy, $E=\frac{1}{2}m{\omega }^2{a}^2$
$E=2{\pi }^{2}m{n}^2{a}^2$
$and{a}^2\propto \frac{1}{m{n}^2}(\because Eissame)$
$\therefore \frac{a_1^2}{a_2^2}=\frac{m_2{n}_2^2}{m_1{n}_1^2}$
Given, $n_2>n_{1}and{m}_1=m_2\Rightarrow a_1>a_2.$
So, amplitude of B smaller than A