The equation of a transverse wave traveling along a very longstring is given by

Question

The equation of a transverse wave traveling along a very longstring is given by y = 6.0sin(0.019x +3.6t), wherex and y are expressed in centimeters andt is in seconds. Determine the following values. b) the wavelength
(c) the frequency d) the speed hz f) the maximum transverse speed of a particle in thestring
g) the transverse displacement at x = 3.5 cm whent = 0.26 s
cm The equation of a transverse wave traveling along a very longstring is given by y = 6.0sin(0.019x +3.6t), wherex and y are expressed in centimeters andt is in seconds. Determine the following values. b) the wavelength
(c) the frequency d) the speed hz f) the maximum transverse speed of a particle in thestring
g) the transverse displacement at x = 3.5 cm whent = 0.26 s
cm

Explanation / Answer

Given   y = 6.0 sin(0.019x + 3.6t), compare this with y = A sin ( kx + w t ) we get amplitude A = 6 cm propagation constant k = 0.019 angular frequency w = 3.6 (b). wavelength = 2 / k                           = 105.26 cm (c). frequency f = w / 2                         = 1.8 Hz (d). speed v = w / k                    = 189.47 cm / s (f) . maximum transverse speed v = A w                                                    =67.85 cm / s (g).  y = 6.0sin(0.019*3.5 +3.6*0.26)          = 0.3296cm

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