Answer:
The equation of a transverse wave traveling along a very long string is y=6.0sin(0.020πx+4.0πt), 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, (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=3.5 cm when t=0.26s?/(a) The amplitude is y
m
=6.0 cm.
(b) We find λ from 2π/λ=0.020π⋅λ=1.0×10
2
cm.
(c) Solving 2πf=ω4.0π, we obtain f=2.0 Hz.
(d) The wave speed is v=λf=(100 cm)(2.0 Hz)=2.0×10
2
cm/.s
(e) The wave propagates in the −x direction, since the argument of the trig function is kx+ωt instead of kx−ωt
(f) The maximum transverse speed (found from the time derivative of y) is
u
max
=2πfy
m
=(4.0πs
−1
)(6.0 cm)=75 cm/s
(g) y(3.5 cm,0.26 s)=(6.0 cm)sin[0.020π(3.5)+4.0π(0.26)]=2.0 cm.