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 '. M 1 = M 2 and L 1 = 2

Question

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 ​ '. M 1 ​ = M 2 ​ and L 1 ​ = 2 L 2 ​ . If their total energies are same then the correct statement is

TestHub · Accepted Answer

The total energy of a simple pendulum executing 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 this into the energy equation, we get

$E = \frac{1}{2} m (\frac{g}{L}) A^{2}$

For pendulum $A$ and pendulum $B$ , the total energies are equal ( $E_{A} = E_{B}$ ). Thus,

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

Given that $M_{1} = M_{2}$ , we can cancel the common terms to obtain

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

Substituting $L_{1} = 2 L_{2}$ into the equation gives

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

$A_{A}^{2} = 2 A_{B}^{2}$

$A_{A} = 2 A_{B}$

This shows that $A_{B} = \frac{A _{A}}{2}$ , which means the linear amplitude of $B$ is smaller than the linear amplitude of $A$ .

Answer: linear amplitude of $B$ is smaller than linear amplitude of $A$ .

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