A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is: mg mg(1 + (A/L) 2 ) mg(1+(A/2L) 2 ) mg(1-(3A/L) 2 )

Question

Arun · Answer

Dear student Tmax = mg + mv&sup2;/l At mean position V = a * w and w = sqrt(g/l) v = a sqrt (g/l) Hence Tmax = mg ( 1 + a&sup2;/l&sup2;) Hence option 2 is correct. Regards Arun (askIITians forum expert)

Rishi Sharma · Answer

Dear Student,
Please find below the solution to your problem.

Tmax = mg + mv²/l
At mean position
V = a * w and w = sqrt(g/l)
v = a sqrt (g/l)
Hence
Tmax = mg ( 1 + a²/l²)

Thanks and Regards

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A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is: mg mg(1 + (A/L) 2 ) mg(1+(A/2L) 2 ) mg(1-(3A/L) 2 )

A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is:
mg
mg(1 + (A/L)²)
mg(1+(A/2L)²)
mg(1-(3A/L)²)

2 Answers

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A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is: mg mg(1 + (A/L) 2 ) mg(1+(A/2L) 2 ) mg(1-(3A/L) 2 )

A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is:mgmg(1 + (A/L)2)mg(1+(A/2L)2)mg(1-(3A/L)2)

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A simple pendulum wiht length L and mass m of bob is vibrating with an amplitude of A. The maximum tension in the string is:
mg
mg(1 + (A/L)²)
mg(1+(A/2L)²)
mg(1-(3A/L)²)