A harmonic wave is a wave that undergoes simple harmonic motion as it travels. It can be described by the equation D(x,t)=Asin(kx-wt+φ), where D is displacement, A is amplitude, k is wave number, w is angular frequency, x is position, t is time, and φ is phase. The document provides examples of how to use this equation to calculate properties like wavelength, frequency, and acceleration of a harmonic wave traveling through different media. It also includes sample problems solving for these wave properties given values for variables like velocity, frequency, amplitude, and position over time.

2. 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

3. The wave travels out the window as seen in the
PhET simulation.

4. D(x) = Asin(kx)
D(x) – displacement of a particle on the string at x
D(x)
x

5. D(x) = Asin(kx)
A – amplitude/positive max displacement of the wave
A
Crest
Trough
D(x) = -A
D(x) = +A

6. D(x) = Asin(kx)
λ- wavelength/distance between crests or trough
λ
λ

7. 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.

8. D(x) = Asin(kx)
k in a harmonic wave is NOT the same as the
k for the spring constant!

9. 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)

10. 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)

11. 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

12. Problem #1 - Answer
B)
The frequency of the wave does not change when
travelling through different mediums.

13. 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?

14. Problem #2 - Answers
i) Find the wavelength, period, and wave number.
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

15. Problem #2 - Answers
i) Find the wavelength, period, and wave number.
The period of the wave is reciprocal of the frequency:
T = 1/ƒ
T = 1/300 = 0.00333s = 3.33ms

16. Problem #2 - Answers
i) Find the wavelength, period, and wave number.
The wave number can be found using this equation:
k = 2π/λ
k = 2π/8.333m = 0.754 rad/m

17. Problem #2 - Answers
ii) Write the wave function in terms of position and
time that is travelling in the direction of increasing x.
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

18. Problem #2 - Answers
ii) Write the wave function in terms of position and time
that is travelling in the direction of increasing x.
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)

19. Problem #2 - Answers
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)

20. Problem #2 - Answers
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

21. 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