harmonic waves

What is a Harmonic wave?

A wave that is undergoing simple harmonic motion.

A travelling wave

While a string oscillates, a harmonic wave travels along

the string

The wave travels out the window as seen in the

PhET simulation.

D(x) = Asin(kx)

D(x) – displacement of a particle on the string at x

D(x)

x

D(x) = Asin(kx)

A – amplitude/positive max displacement of the wave

A

Crest

TroughD(x) = -A

D(x) = +A

D(x) = Asin(kx)

λ- wavelength/distance between crests or trough

λ

λ

D(x) = Asin(kx)

k – the wave number

frequency of wave pattern per metre

Change of phase per unit length

k = 2π/λ

We relate k with λ with the above equation based on a sine graph

k = 1 if the wavelength is equal to 2π

k is inversely proportional to λ(with constant 2π)

Note: If we use the same graph but change the axis to time(t) instead of

position(x), k = 1 when the period is equal to 2π.

D(x) = Asin(kx)

k in a harmonic wave is NOT the same as the

k for the spring constant!

D(x, t) = Asin((2π/λ)x ± (2π/T)t)

This equation describes a 3D plot of a travelling harmonic wave

From the previous function, we replace x by x – vt when the

wave is travelling towards positive x (right)

We replace x by x + vt when the wave is travelling towards

negative x (left)

D(x, t) = Asin((2π/λ)x ± (2π/T)t + 𝝓)

Velocity(v) = λƒ = w/k

So, D(x, t) = Asin((2π/λ)x ± (2π/T)t)

To generalize the equation: we add 𝝓

D(x, t) = Asin((2π/λ)x ± (2π/T)t + 𝝓)

Problem #1

The speed of sound waves in oil is 1461m/s and

1400m/s in water. The frequency of the sound wave is

550 Hz as it travels through the water. What is the

frequency of the sound wave as it travels from water to

oil?

A) <550Hz B) 550Hz C) >550Hz

D) Not enough information

Problem #1 - Answer

B)

The frequency of the wave does not change when

travelling through different mediums.

Problem #2

A harmonic wave is travelling through molten tin at 2500 m/s with a frequency of 300Hz. The maximum displacement is 5mm.

i) Find the wavelength, period, and wave number.

ii) Write the wave function in terms of position and time that is travelling in the direction of increasing x.

iii) What is the acceleration at 0.001s of a segment of the wave located at 7.0 nm?

Problem #2 - Answers


Since we know the velocity and the frequency of the

wave, we can use this equation:

Velocity(v) = λƒ

Solving for λ, we get λ=v/ƒ

λ= 2500m/s / 300 Hz = 8.333m



The period of the wave is reciprocal of the frequency:

T = 1/ƒ

T = 1/300 = 0.00333s = 3.33ms



The wave number can be found using this equation:

k = 2π/λ

k = 2π/8.333m = 0.754 rad/m



The general formula is: D(x, t) = Asin((2π/λ)x ±(2π/T)t + 𝝓)

The maximum displacement is the amplitude of the wave.

A = 0.005m



Since no phase constant is given we assume 𝝓 = 0

It is travelling in the direction of increasing x and so we subtract the component in relation to time.

Therefore, the wave function is:

D(x, t) = 0.005sin(0.754x - 1887t)


iii) What is the acceleration at 0.001s of a segment of

the wave located at 7.0 nm?

Taking the partial derivative of the general wave

function twice with respect to time, we end up with

this equation:

a(x, t) = -w2Asin(kx – wt + 𝝓)


iii) What is the acceleration at 0.001s of a segment of

the wave located at 7.0 nm?

a(x , t) = -w2Asin(kx – wt + 𝝓)

w = 1887rad A = 0.005m k = 0.754rad/m

x = 7 x 10-9m t = 0.001s 𝝓 = 0

a(x, t) = 16921m/s2

Works Cited

Wave on a String PhET simulation (images)

http://phet.colorado.edu/en/simulation/wave-on-a-string

Physics 101 Textbook (definitions and equations)

Physics for Scientists and Engineers – An Interactive

Approach

harmonic waves

Science