InterviewSolution
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InterviewSolution
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A string is fixed at both end transverse oscillations with amplitude `a_(0)` are excited. Which of the following statements are correct ?A. (a) Energy of oscillations in the string is directly proportional to tension in the stringB. (b) Energy of oscillations in nth overtone will be equal to `n^(2)` times of that in first overtoneC. ( c) Average kinetic energy of string (over an oscillation period) is half of the oscillation energyD. (d) None of the above |
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Answer» Correct Answer - A::C If a string of length `l` has cross - sectional area `A`, density of its material `rho` then its oscillation energy is given by `E = pi^(2) A rho a_(0)^(2) l f^(2)` where `f` is frequency of transverse stationary wave formed in the string . But `f = (v)/(lambda) = (1)/(lambda) sqrt((T)/(m))` where `lambda` is wavelength , `T` is tension in the string and `m = A rho`. Since , string has a fixed length , therefore , wavelength of a tone excited in the string is constant . Hence , energy `E prop T`. Therefore , option `(a)` is correct . If the frequency of fundamental tone is `f_(0)`, then the frequency on `n th ` overtone will be equal to `( n + 1) f_(0)`. Hence , oscillation energy of the string will be equal to : ` E_(n) = pi^(2) A rho a_(0)^(2) l f_(0)^(2) ( n + 1)^(2)` Since `E_(n)` is not directly proportional to `n^(2)` , therefore , option `(b)` is wrong . Since , every particle of the string performs `SHM`, therefore , `r.m.s.` speed of a particle ` = 1// sqrt(2) xx` its maximum speed Hence , average `KE` is half of maximum `KE`. But maximum `KE` is equal to oscillation energy of the string . therefore , option `(c )` is correct. |
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