Two simple pendulums, A of mass M 1 and length L 1 , and B of mass M 2 and length L 2 . M 1 = M 2 and L 1 = 2 L 2 . If vibrational energies of both are same then which one of the follo

Question

Two simple pendulums, A of mass M 1 ​ and length L 1 ​ , and B of mass M 2 ​ and length L 2 ​ . M 1 ​ = M 2 ​ and L 1 ​ = 2 L 2 ​ . If vibrational energies of both are same then which one of the following is correct?

TestHub · Accepted Answer

The vibrational energy of a simple pendulum in simple harmonic motion is given by $E = \frac{1}{2} m ω^{2} A^{2}$ .

For a simple pendulum, the angular frequency is $ω = \frac{g}{L}$ .

Substituting $ω$ into the energy equation, we get $E = \frac{1}{2} m (\frac{g}{L}) A^{2}$ .

Given that the vibrational energies are equal ( $E_{1} = E_{2}$ ) and the masses are equal ( $M_{1} = M_{2}$ ):

$\frac{1}{2} M_{1} \frac{g}{L _{1}} A_{1}^{2} = \frac{1}{2} M_{2} \frac{g}{L _{2}} A_{2}^{2}$

$\frac{A _{1}^{2}}{L _{1}} = \frac{A _{2}^{2}}{L _{2}}$

Since $L_{1} = 2 L_{2}$ , we can substitute this into the equation:

$\frac{A _{1}^{2}}{2 L _{2}} = \frac{A _{2}^{2}}{L _{2}}$

$A_{1}^{2} = 2 A_{2}^{2} \Rightarrow A_{1} = 2 A_{2}$

Thus, $A_{1} > A_{2}$ , which means the amplitude of pendulum B is smaller than that of pendulum A.

Answer: Amplitude of B is smaller than A

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