PHYSIC 難題10

I suppose the string is vibrating at a frequency of 190 Hz, which is 5 times the fundamental frequenct. Hence, the stationary wave should contain 5 loops. Point X is thus at the mid-point of the 3rd loop (i.e. the centre loop).

Period of oscillation = 1/190 s = 5.263 x 10^-3 s = 5.263 ms

(1) At t = 0 ms, point X is at maximum displacement. Since in a stationary wave, all points are vibrating in phase. Hence, the wave shows 5 loops with max displacement of the anti-nodes.

(2) 10.53 ms = 10.53/5.263 periods = 2 periods
The wave form is the same as at t = 0 ms.

(3) 18.42 ms = 18.42/5.263 periods = 3.5 periods
In 3 complete periods, the wave form remains the same as at t=0 ms. After a further of 0.5 period, point X should be at the lower max displacement. Hence, the entire wave form is inverted.

(4) 22.37 ms = 22.37/5.263 period = 4.25 periods
In 4 complete periods, the wave form appears as what it is at t = 0 ms. A further lapse of 0.25 (i.e. 1/4) period would make point X be back to the equilibrium position. Since all points in a stationary wave are in phase, the wave form at that instant of time is a straight line.