Question

The equation of a transverse wave traveling along a very long string is y= 4.61 sin(0.0599πχ+ 9.2От), where x and y are expressed in centimeters and tis in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x= 602 cm when t= 0.343 s? Unitsl cm (a) NumberT4.61 (b) NumbehT 33.39 (c) NumbetT4.6 (d) Number (e) 1 (f) Numbe Units cm/s Units (rad's or 1 Units Units (g) Numbe Units Click if you would like to Show Work for this question: Open Show Work

Answer 1

b) unit will be cm.

c) here the unit will be Hz

d) the speed of a wave is given by

v=f$\lambda$=4.6*33.39=153.594cm/s

so answer is 153.59 cm/s or 153.6 cm/s or 154 cm/s

e) answer=negative x-axis

because $\omega$ is positive in the following wave equation

f) the maximum transverse speed is given by

vm=$\omega$*amplitude=9.2pi*4.61=133.174cm/s

so the answer is 133.17 cm/s or 133.2 cm/s or 133 cm/s

g) the wave equation is

y=4.61sin(0.0599pix+9.20pit)

substituting the value of x and t we get

y=4.61 sin( 0.0599pi*6.02+9.2pi*0.343)=-4.61

so the answer is -4.61cm