First slide

Applications of SHM

Question

Two simple pendulum first of bob mass M₁ and length L₁ second of bob mass M₂ and length L₂. M₁= M₂and L₁=2L₂.If these vibrational energy of both is same. Then which is correct

Difficult

A

Amplitude of B greater than A
B

Amplitude of B smaller than A
C

Amplitudes will be same
D

None of these

Solution

$\begin{array}{l} n = \frac{1}{2 π} \sqrt{\frac{g}{l}} \Rightarrow n \propto \frac{1}{\sqrt{l}} \Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{\frac{l_{2}}{l_{1}}} = \sqrt{\frac{L_{2}}{2 L_{2}}} \\ \Rightarrow \frac{n_{1}}{n_{2}} = \frac{1}{\sqrt{2}} \Rightarrow n_{2} = \sqrt{2} n_{1} \Rightarrow n_{2} > n_{1} \\ Energy, E = \frac{1}{2} m ω^{2} a^{2} = 2 π^{2} m n^{2} a^{2} \\ \Rightarrow \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{m_{2} n_{2}^{2}}{m_{1} n_{1}^{2}} (∵ E is same) \end{array}$
$Given n_{2} > n_{1} and m_{1} = m_{2} \Rightarrow a_{1} > a_{2}$

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