Two simple pendulums have first (A) bob of mass ' M₁ ' and length ' L₁ ', second (B) of mass ' M₂ ' and length ' L₂ '. M₁ = M₂ and L₁ = 2L₂. If their total energies are same then the correct statement is (a) amplitude of B is greater than amplitude of A. (b) amplitude of B is smaller than amplitude of A. (c) amplitude of both will be same. (d) amplitude of B is twice that of A.

Answer: (b). Energy of a pendulum with linear amplitude A: E = 1/2mω²A² = 1/2mg/LA². Same E and same m ⇒ A² ∝ L. L₁ = 2L₂ ⇒ AA² = 2AB², so AB = AA/√(2). So the answer is (b) amplitude of B is smaller than amplitude of A.

Two simple pendulums have first (A) bob of mass ' M₁ ' and length '… | MHT CET 2025 (25 Apr, Shift 2)

Q27MHT CET 2025 · 25 APR · SHIFT 2Medium

Simple Pendulum

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 $L_1 = 2L_2$ . If their total energies are same then the correct statement is

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Answer: (b)

Energy of a pendulum with linear amplitude $A$ : $E = \frac{1}{2}m\omega^2A^2 = \frac{1}{2}m\frac{g}{L}A^2$ .
Same $E$ and same $m \Rightarrow A^2 \propto L$ .
$L_1 = 2L_2 \Rightarrow A_A^2 = 2A_B^2$ , so $A_B = A_A/\sqrt{2}$ .
So the answer is (b) amplitude of B is smaller than amplitude of A.

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Simple Pendulum question from MHT CET 2025 (25 Apr, Shift 2)

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