Two simple pendulum first of bob mass M_1 and length L_1 second o… - Vidyadip

MCQ

Two simple pendulum 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 these vibrational energy of both is same. Then which is correct

A
Amplitude of $B$ greater than $A$
✓
Amplitude of $B$ smaller than $A$
C
Amplitudes will be same
D
None of these

✓

Answer

Correct option: B.

Amplitude of $B$ smaller than $A$

b
(b) $n = \frac{1}{{2\pi }}\sqrt {\frac{g}{l}} $==> $n \propto \frac{1}{{\sqrt l }}$==> $\frac{{{n_1}}}{{{n_2}}} = \sqrt {\frac{{{l_2}}}{{{l_1}}}} = \sqrt {\frac{{{L_2}}}{{2{L_2}}}} $

==> $\frac{{{n_1}}}{{{n_2}}} = \frac{1}{{\sqrt 2 }}$==> ${n_2} = \sqrt 2 \,{n_1}$ ==> ${n_2} > {n_1}$

Energy $E = \frac{1}{2}m{\omega ^2}{a^2} = 2{\pi ^2}m{n^2}{a^2}$

==>$\frac{{a_1^2}}{{a_2^2}} = \frac{{{m_2}n_2^2}}{{{m_1}n_1^2}}$ ( $E$ is same)

Given ${n_2} > {n_1}$ and ${m_1} = {m_2}$ ==> ${a_1} > {a_2}$

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