To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y (x, t) = Acos (kx-ωt). A transverse wave on a string is traveling in the + x direction with a wave speed of 8.50 m/s, an amplitude of 5.50*10-2 m, and a wavelength of 0.500 m. At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.52 m and t = 0.150 s.

0.0549 m

Explanation:

Given that

equation y (x, t) = Acos (kx-ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^-2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π / λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~ = 107 rad/sec

so

equation y (x, t) = Acos (kx-ωt)

y (x, t) = 5.5*10^-2 cos (12.568 x-107t)

when x = 0 and and t = 0

maximum y (x, t) = 5.5*10^-2 cos (12.568 (0) - 107 (0))

maximum y (x, t) = 5.5*10^-2 m

and when x = x = 1.52 m and t = 0.150 s

y (x, t) = 5.5*10^-2 cos (12.568 (1.52) - 107 (0.150))

y (x, t) = 5.5*10^-2 * (0.9986)

y (x, t) = 0.0549 m

so the transverse displacement is 0.0549 m