Question 229: Oscillations

Question

Two simple pendulums have first (A) bob of mass ' $M_1$ ' and length ' $L_1$ ', second (B) of mass ' $\mathrm{M}_2$ ' and length ' $\mathrm{L}_2$ '. $\mathrm{M}_1=\mathrm{M}_2$ and $\mathrm{L}_1=2 \mathrm{~L}_2$. If their total energies are same then the correct statement is

Accepted Answer

Total energy of a simple pendulum,
$$ E=\frac{1}{2} M \omega^2 A^2 $$
For a simple pendulum,
$$ \omega=\sqrt{g / L} $$
For pendulum A:
$$ E_A=\frac{1}{2} M_1\left(\frac{g}{L}ight) A_1^2 $$
For pendulum B :
$$ E_B=\frac{1}{2} M_2\left(\frac{g}{L}ight) A_2^2 $$
Given:
$$ \begin{array}{ll}  & E_A=E_B \ & \frac{1}{2} M_1\left(\frac{g}{L_1}ight) A_1^2=\frac{1}{2} M_2\left(\frac{g}{L_2}ight) A_2^2 \ 	herefore \quad & \frac{A_1^2}{A_2^2}=\frac{L_1}{L_2} \quad \ldots \ldots .\left(M_1=M_2ight) \end{array} $$
$$ 	herefore \quad \mathrm{A}_1^2=2 \mathrm{~A}_2^2 \quad \ldots \ldots\left(\mathrm{~L}_1=2 \mathrm{~L}_2ight) $$
$$ \begin{aligned} &	herefore \quad \mathrm{A}_1>\mathrm{A}_2\ &	herefore \quad 	ext { Amplitude of } B 	ext { is smaller than amplitude of } A \end{aligned} $$

Answer

amplitude of B is greater than amplitude of $A$.

Answer

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

Answer

amplitude of both will be same.

Answer

amplitude of B is twice that of A .

Oscillations

Two simple pendulums have first (A) bob of mass ' M_1 ' and length ' L_1 ', second (B) of mass ' \mathrm{M}_2 ' and length ' \mathrm{L}_2 '. \mathrm{M}_1=\mathrm{M}_2 and \mathrm{L}_1=2 \mathrm{~L}_2. If their total energies are same then the correct statement is

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