The equation of a transverse wave traveling along a very long string is y = 4.89

Question

The equation of a transverse wave traveling along a very long string is y = 4.89 sin(0.0873pix+ 2.50pit), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, and (e) the maximum transverse speed of a particle in the string. (f) What is the transverse displacement at x = 2.44 cm when t = 0.818 s? Give your answers in centimeter-based units, where applicable. (a) Number 4.89 Units cm (b) Number 2290 Units cm (c) Number 1.25 Units Hz (d) Number 2863 Units cm/s (e) Number 3840 Units cm/s (f) Number 0.604 Units cm

Explanation / Answer

A) amplitude is 4.89 cm

B) wavelength lamda = (2*pi/k) = (2*pi/0.0873*pi) = 22.9 cm....

C) f= w/(2*pi) = 2.5*pi/(2*pi) = 1.25 Hz...

D) speed v = w/k = 2.5/0.0873 = 28.63 cm/sec...

e) Vmax = Aw = 4.89*2.5*3.142 = 38.41 cm/sec..

f) y = 4.89 sin[(0.0873*3.142*2.44)+(2.5*3.142*0.818)]....

y = 3.546 cm

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