If two simple pendulums first of bob mass M1 and length l1, second of bob mass M2 and length l2. Given M1=M2 and l1=2l2. If the vibrational energies of both are same, then which of the following is correct?
A
Amplitude of B is smaller than A
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B
Amplitude of B is greater than B
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C
Amplitude will be same
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D
None of the above
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Solution
The correct option is C Amplitude of B is smaller than A Frequency, n=12π√gl or n∝1√l ∴n1n2=√l2l1=√l22l2 n1n2=1√2 ⇒n2=√2n1 ⇒n2>n1 Energy, E=12mω2a2 and a2∝1mn2 {∵E is same} ∴a21a22=m2n22m1n21 Given, n2>n1 and m1>m2 ⇒a1>a2 So, amplitude of B is smaller than A.