A string of length 'l' is fixed at both ends. It is vibrating in tis 3rd overtone with maximum amplitude 'a' the amplitude at a distance `l/3` from one end is

Options

A. a
B. 0
C. `(sqrt(3)a)/2`
D. `a/2`

Answer

C

Explanation

For a string vibrating in its `n^(th)` overtone (`(n+1)^(th)` harmonic)
`y=2Asin (((n+1)pi x)/(L)) cos omegat`

For `x=l/3, 2A=a` and n=3, `y=[asin((4pi)/l.l/3)] cos omegat`
`=a sin .(4pi)/3 cos omegat=a((sqrt(3))/2) cos omegat`
i.e., at `x=l/3` the amplitude is `(sqrt(3)a)/2`