A string is fixed at both end transverse oscillations with amplitude `a_(0)` are excited. Which of the following statements are correct ?

Options

A. (a) Energy of oscillations in the string is directly proportional to tension in the string
B. (b) Energy of oscillations in nth overtone will be equal to `n^(2)` times of that in first overtone
C. ( c) Average kinetic energy of string (over an oscillation period) is half of the oscillation energy
D. (d) None of the above

Answer

A, C

Explanation

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.