The equation of a transverse wave traveling along a very long string is y = 6.0 sin (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 = 0.26 s?

a) A = 6.0 cm, b) λ = 100 cm, c) f = 0.5 Hz, d) v = 0.5 m / s, e) the wave propagates to the right, f) v = 0.12π m / s, g) y = - 1.67 cm

Explanation:

The general equation of a wave on a string going off to the right is

y = A sin (kx - wt)

where A is the amplitude of the movement. w the angular velocity and k the wave number

in this exercise the equation remain is

y = 6.0 sim (0.020π x - 4.0π t)

a) Let's compare the two equations to find the answers

the amplitude is

A = 6.0 cm

b) The wave number is

k = 2π/λ

λ = 2π / k

λ = 2π / 0.020π

λ = 100 cm

c) angular velocity is related to frequency

w = 2π f

f = 2π / w

f = 2π / 4.0π

f = 0.5 Hz

d) at wave speed is given by

v = λ f

v = 1 0.5

v = 0.5 m / s

e) as the temporal part is negative, the wave propagates to the right

f) we find the maximum transverse speed from the equation

v = dy / dt

v = - A w cos kx-wt)

the speed is maximum when the cosine function is ±1

v = - A w

v = 6.0 0.020π

v = 0.12π m / s

g) the displacement for x = 3.5 cm t = 0.26 s

y = 6.0 sin (0.020π 6.5 - 4.0π 0.26)

y = 6.0 (-0.279)

y = - 1.67 cm