Question

For two simple pendulums first has bob mass $M_{1}$ and length $l_{1}$ , second has bob mass $M_{2}$ and length $l_{2}$ . Given $M_{1}=M_{2} \, and \, l_{1}=2l_{2}$ . If the vibrational energies of both are same, then which of the following is correct?

• A. Amplitude of $B$ is smaller than $A$
• B. Amplitude of $B$ is greater than $B$
• C. Amplitude will be same
• D. None of the above

Answer 1

Answer: (A)

Frequency, $n=\frac{1}{2 \pi }\sqrt{\frac{g}{l}}$
Or $n \propto \frac{1}{\sqrt{l}}$
$\therefore $ $ \, \frac{n_{1}}{n_{2}}=\sqrt{\frac{l_{2}}{l_{1}}}=\sqrt{\frac{l_{2}}{2 l_{2}}}$
$\frac{n_{1}}{n_{2}}=\frac{1}{\sqrt{2}}$
$\Rightarrow \, \, \, $ $n_{2}=\sqrt{2 n_{1}}$
$\Rightarrow \, \, \, $ $n_{2}>n_{1}$
Energy, $E=\frac{1}{2}m\omega ^{2}a^{2}$
And $a^{2} \propto \frac{1}{m n^{2}}$ $\left\{\because \, \, E \, i s \, s a m e\right\}$
$\therefore \, \, $ $\frac{a_{1}^{2}}{a_{2}^{1}}=\frac{m_{2} n_{2}^{2}}{m_{1} n_{1}^{2}}$
Given, $n_{2}>n_{1} \, and \, m_{1}=m_{2}$
$\Rightarrow \, $ $a_{1}>a_{2}$
So, amplitude of $B$ is smaller than $A$