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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 kinetic at time t`t`, where t is:A. `(1)/(600)sec`B. `(5)/(600)sec`C. `(19)/(600)sec`D. `(25)/(600)sec` |
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Answer» Correct Answer - 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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