Question

Two simple pendulums first of bob mass $M_1$ and length $L_1$, second of bob mass $M_2$ and length $L_2 .M_2 = M_2 and L_1 = 2 L_2.$ 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, $\, \, \, \, \, \, \, \, \, \, n=\frac{1}{2\pi}\sqrt{\frac{g}{l}}$
$or \, \, \, \, \, \, \, \, \, \, n\propto=\frac{1}{\sqrt{l}}$
$\therefore \, \, \, \, \, \, \, \, \, \, \frac{n_1}{n_2}=\sqrt{\frac{l_2}{l_1}}=\sqrt{\frac{L_2}{2L_2}}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{n_1}{n_2}=\frac{1}{\sqrt{2}}$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n_2=\sqrt{2}n_1$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, n_2 > n_1$
Energy, $\, \, \, \, \, \, \, \, \, \, E=\frac{1}{2}m\omega^2 a^2$
$\, \, \, \, \, \, \, \, \, \, E=2\pi^2mn^2 a^2$
$and\, \, \, \, \, \, \, \, \, \, a^2\propto\frac{1}{mn^2}\, \, \, \, \, \, \, \, \, \, \, \, (\because E\, is\, same)$
$\therefore \, \, \, \, \, \, \, \, \, \, \frac{a^2_1}{a^2_2}=\frac{m_2n^2_2}{m_1n^2_1}$
Given, $n_2 > n_1 \,and \,m_1 = m_2 \Rightarrow a_1 > a_2. $
So, amplitude of B smaller than A