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(B) The total energy $E$ of a simple pendulum performing simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\o

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Amplitude of $B$ is smaller than amplitude of $A$.

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(B) The total energy $E$ of a simple pendulum performing simple harmonic motion is given by $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.For a simple pendulum,the angular frequency is $\omega = \sqrt{\frac{g}{L}}$,so $\omega^2 = \frac{g}{L}$.Substituting this into the energy equation: $E = \frac{1}{2} m \left(\frac{g}{L}\right) A^2$.Given $E_A = E_B$,$M_1 = M_2 = M$,and $L_1 = 2 L_2$:$\frac{1}{2} M \left(\frac{g}{L_1}\right) A_A^2 = \frac{1}{2} M \left(\frac{g}{L_2}\right) A_B^2$.Simplifying,we get $\frac{A_A^2}{L_1} = \frac{A_B^2}{L_2}$.Substituting $L_1 = 2 L_2$: $\frac{A_A^2}{2 L_2} = \frac{A_B^2}{L_2}$.This leads to $A_A^2 = 2 A_B^2$,or $A_A = \sqrt{2} A_B$.Therefore,$A_B = \frac{A_A}{\sqrt{2}}$,which means the amplitude of $B$ is smaller than the amplitude of $A$.

Two simple pendulums have first $(A)$ bob of mass $M_1$ and length $L_1$,second $(B)$ of mass $M_2$ and length $L_2$. Given $M_1 = M_2$ and $L_1 = 2 L_2$. If their total energies are the same,then which of the following statements is correct?

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