Question

If two simple pendulums first of bob mass $M_1$ and length $l_1$, second of 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)

Amplitude of $B$ is smaller than $A$

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