A transverse mechanical wave is traveling along a string lying along the x-axis.

Question

A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, y(x,t), is described by the following equation:
y(x,t)=0.0210*sin(3.40x-128t),
where x and y are in meters and the time is in seconds.

a)What is the wavelength of the wave?

b)What is the velocity of the wave? (Define positive velocity along the positive x-axis.)

c)What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)

Explanation / Answer

This function
(1) y(x,t)=0.0350×sin(6.60x−126t)

is a sine wave of the form:
(2) y(x, t) = A * sin(k * x + ω * t)
where
k = wave number in rads / m
ω = angular velocity in rads / s

Equations for a) and b) are taken from Reference 1 on sine waves.

a) What is the wavelength of the wave?

(3) k = 2 * π / λ

Solving for λ:
(4) λ = 2 * π / k = 2 * 3.14 / 6.60 = 0.952m <<<===a) Answer


b) What is the velocity of the wave? (Define positive velocity along the positive x-axis.)

(5) k = ω / c

Solving for c, the phase velocity:
(6) c = ω / k = -126 / 6.60 = -19.1m/s <<<===b) Answer


c) What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)

y(x,t) is the position function, so dy/dt gives the velocity function:

(7) y'(x,t)= v(x, t) = 0.0350 * (-126) * cos (6.60x−126t)

Since we're looking for the maximum speed, and the max value for cos is 1, we have:

(8) vmax = | 0.0350 * (-126) * 1 | = 4.41m/s <<<===c) Answer

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