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Two simple pe...

Question

Two simple pendulum first of bob mass $M_{1}$ and length $L_{1}$ second od bob mass $M_{2}$ and length $L_{2}$ ( $M_{1} = M_{2} and L_{1} = 2 L_{2}$ ).If these vebrational energy of both is same. Then which correct

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Solution

Given,
$L_{1} = 2 L_{2} M_{1} = M_{2}$
As we know,
$n = \frac{1}{2 π} \sqrt{\frac{g}{l}} \Rightarrow n \propto \frac{1}{\sqrt{l}} \Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{l_{2}}{2 l_{1}}} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{\sqrt{2}} \Rightarrow n_{2} = \sqrt{2} n_{1} \Rightarrow n_{2} > n_{1}$
Energy,
$E = \frac{1}{2} m ω^{2} a^{2} = 2 π^{2} m n^{2} a^{2}$
If $E$ is same
$\frac{a_{1}^{2}}{a_{2}^{2}} = \frac{m_{2} n_{2}^{2}}{m_{1} n_{1}^{2}} ∴ n_{2} > n_{1} and m_{1} = m_{2} a_{1} > a_{2}$

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