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