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(B) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$,which implies $n \propto \frac{1}{\sqrt{L}}$.For the two pen

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Amplitude of $B$ is smaller than $A$

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(B) The frequency of a simple pendulum is given by $n = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$,which implies $n \propto \frac{1}{\sqrt{L}}$.For the two pendulums,the ratio of frequencies is $\frac{n_1}{n_2} = \sqrt{\frac{L_2}{L_1}} = \sqrt{\frac{L_2}{2L_2}} = \frac{1}{\sqrt{2}}$.Thus,$n_2 = \sqrt{2}n_1$,which means $n_2 > n_1$.The vibrational energy of a simple harmonic oscillator is $E = 2\pi^2 m n^2 a^2$,where $a$ is the amplitude.Since $E_1 = E_2$,$m_1 = m_2$,and $n_2 > n_1$,we have $n_1^2 a_1^2 = n_2^2 a_2^2$.Therefore,$\frac{a_1^2}{a_2^2} = \frac{n_2^2}{n_1^2} = 2$,which implies $a_1 = \sqrt{2}a_2$.Since $a_1 > a_2$,the amplitude of pendulum $B$ is smaller than that of pendulum $A$.

Two simple pendulums are given: the first has a bob of mass $M_1$ and length $L_1$,and the second has a bob of mass $M_2$ and length $L_2$. Given $M_1 = M_2$ and $L_1 = 2L_2$. If the vibrational energy of both is the same,which of the following statements is correct?

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