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The equation for the displacement of a stretched string is given by y = 4sin 2pi [t/0.02 - x/100] where y and x are in cm and t in sec. Determine the (a) direction in which wave is propagating (b) amplitude (c) time period (d) frequency (e) angular frequency (t) wavelength (g) propagation constant (h) velocity ofwave (i) phase constant and (j) the maximum particle velocity. |
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Answer» Solution :Comparing the GIVEN, equation with the general wave equation : `y = A sin (omegat - Kx = phi)`, i.e., `y = A sin 2pi [t/T-x/lambda+phi]` we find that : (a) As there is negative sign between t and x TENNS, the wave is propagating along positive x-axis. (b) The amplitude of the wave A= 4 cm = 0.04m. (c) The time period ofthe wave T= 0.02 s = (1150) s. (d) The frequency of the wave f = (1 /T) =50Hz. (e) Angularfrequency ofthe wave `OMEGA =2pi f= 100 pi " red"//s` (t) The wave length of the wave `lambda =100cm = 1m` . (g) The propagation CONSTANT. (= wave vector = angular wave number `=(2pi//lambda) =2pi "red/m".` (h) The velocity of wave `v = flambda =50 xx 1 =50 m//s` . (i) The phase constant. i.e., initial phase `phi = 0`. (j) The maximum particle velocity `(V_(Pa))_(max) =A omega = 0.04 xx 100pi = 4pim//s` |
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