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MHT CET · Physics · Oscillations

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

  1. A amplitude of \(B\) is greater than amplitude of \(A\).
  2. B amplitude of \(B\) is smaller than amplitude of \(A\).
  3. C amplitude of both will be same.
  4. D amplitude of \(B\) is twice that of \(A\).
Verified Solution

Answer & Solution

Correct Answer

(B) amplitude of \(B\) is smaller than amplitude of \(A\).

Step-by-step Solution

Detailed explanation

\(E = \frac{1}{2} \frac{mg}{L} A^2\) \(E_1 = E_2 \implies \frac{1}{2} \frac{M_1 g}{L_1} A_1^2 = \frac{1}{2} \frac{M_2 g}{L_2} A_2^2\)