The general equation describing a wave is:

y(x,t) = y_max sin(kx - ωt)

Let's say that for a particular wave on a string the equation is:

y(x,t) = (0.9 cm) sin[(1.2 m^-1)x - (5.0 s^-1)t]

(a) Determine the wave's amplitude, wavelength, and frequency.

(b) Determine the speed of the wave.

(c) If the string has a mass/unit length of μ = 0.012 kg/m, determine the tension in the string.

(d) Determine the direction of propagation of the wave.

(e) Determine the maximum transverse speed of the string.

Solutions

Part (a): The wave's amplitude, wavelength, and frequency can be determined from the equation of the wave:

y(x,t) = (0.9 cm) sin[(1.2 m^-1)x - (5.0 s^-1)t]

The amplitude is whatever is multiplying the sine.
A = y_max = 0.9 cm

The wavenumber k is whatever is multiplying the x:
k = 1.2 m^-1

The wavelength is

λ

=

2π k

= 5.2 m

The angular frequency ω is whatever is multiplying the t.
ω = 5.0 rad/s

f

=

ω 2π

= 0.80 Hz

Part (b): The wave speed can be found from the frequency and wavelength:

v = f λ = 0.80 * 5.2 = 4.17 m/s

Part (c): With μ = 0.012 kg/m and the wave speed given by:

v

=

(

T μ

)

½

This gives a tension of T = μ v² = 0.012 (4.17)² = 0.21 N.

Part (d): To find the direction of propogation of the wave, just look at the sign between the x and t terms in the equation. In our case we have a minus sign:

y(x,t) = (0.9 cm) sin[(1.2 m^-1)x - (5.0 s^-1)t]

A negative sign means the wave is traveling in the +x direction.

A positive sign means the wave is traveling in the -x direction.

Part (e): To determine the maximum transverse speed of the string, remember that all parts of the string are experiencing simple harmonic motion. We showed that in SHM the maximum speed is:

v_max = Aω

In this case we have A = 0.9 cm and ω = 5.0 rad/s, so:

v_max = 0.9 * 5.0 = 4.5 cm/s

For any one piece of the medium, the kinetic energy is a maximum when the piece passes through the equilibrium position. The potential energy is also maximum there - that's where the springs connecting the piece to its neighbors are stretched the most. The average kinetic energy equals the average potential energy.

The forces between the pieces of the medium continually do work to transport energy through the medium. The energy being transported is the energy being put into the system by whatever is shaking the first piece in the medium.

The book derives the fact that the average power being transmitted by a wave on a string is:

P_avg = ½ μ ω ² A² v

Let's return to our previous sample problem, with a wave on a string given by:

y(x,t) = (0.9 cm) sin[(1.2 m^-1)x - (5.0 s^-1)t]

The mass/unit length of the string is μ = 0.012 kg/m, and we determined that:

v = 4.17 m/s

What is the average power transmitted by the wave?

P_avg = ½ μ ω ² A² v

P_avg = ½ (0.012) (4.17) (5.0)² (0.009)² = 5.1 x 10^-5 W