Two simple pendulum first of bob mass $$M1$$ and length $$L1$$ second of bob mass $$M2$$ and length $$L2$$. $$M1=$$ $$M2$$ and $$L1=2L2$$.If these vibrational energy of both is same. Then which is correct

Correct option is 2Amplitude of B smaller than A

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Two simple pendulum first of bob mass M₁ and length L₁ second of bob mass M₂ and length L₂. M₁= M₂and L₁=2L₂.If these vibrational energy of both is same. Then which is correct

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Amplitude of B greater than A

Amplitude of B smaller than A

Amplitudes will be same

None of these

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detailed solution

Correct option is B

n=12πgl⇒n∝1l⇒n1n2=l2l1=L22L2⇒ n1n2=12⇒n2=2n1⇒n2>n1 Energy, E=12mω2a2=2π2mn2a2⇒ a12a22=m2n22m1n12 (∵E is same ) Given n2>n1 and m1=m2⇒a1>a2

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Two simple pendulum first of bob mass M1 and length L1 second of bob mass M2 and length L2. M1= M2 and L1=2L2.If these vibrational energy of both is same. Then which is correct

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Two simple pendulum first of bob mass M₁ and length L₁ second of bob mass M₂ and length L₂. M₁= M₂and L₁=2L₂.If these vibrational energy of both is same. Then which is correct