Saved Bookmarks
| 1. |
A simple harmonic wave of amplitude 0.2 cm, frequency 1000 Hz and wavelength 0.31 m is travelling along the positivex-axis . Calculate the displacement of the particle at 3.1 m from the origin after 1.004 s. What would be the phase difference for two positions of the vibrating particle after an interval of 0.001 s ? |
|
Answer» SOLUTION :Data : A= 0.2 cm = 0.002 m, n = 1000 Hz, ` lamda = 0.31m ,x = 3.1 m` ` t = 1.004 s, t_(2) -t_(1) = 0.001 s` (i) The displacement of the particle. ` y = A sin 2 pi (NT -x/lamda)` ` = 0.002 sin 2 pi(1000 xx 1.004 - (3.1)/(0.31))` ` = 0.002 sin 2 pi (1004 -10)` ` = 0.002 sin 2pi (994)` ` y = 0 [ or y =0` metre] `2 pi xx 1000 xx 0.001 = 2pi` radians |
|