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A string fixed at both the ends of lengt...

A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by `y=Asinkxcos(omegat+(pi)/(3))`. All the symbols have their usual meaning. Mass per unit length of the string is `0.5(gm)/(cm)`. Given that `A=1` cm and `omega=100pi(rad)/(sec)` Answer the following 2 questions based on information given (use `mu^(2)=10)`
Q. Starting from `t=0`, energy of vibration is completely potential at time `t`, where t is

A

`(1)/(600)sec`

B

`(5)/(600)sec`

C

`(19)/(600)sec`

D

`(25)/(600)sec`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
For energy to be completely potential
`cos(omegat+(pi)/(3))=+-1impliesomegat+(pi)/(3)=npi`
`t=((3n-1)/(3))(pi)/(omega)=((3n-1)/(300))sec`.
For energy to be completely kinetic
`cos(omegat+(pi)/(3))=0`
`omegat+(pi)/(3)=(2n-1)(pi)/(2)`
`t=((6n-5)/(600))sec`
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