Drawing a wave from its equation

From the Waves chapter to code

In the Waves chapter you wrote a travelling wave as an equation. Here you draw it with numpy and watch it move.

A wave on a rope has a displacement $y$ at every position $x$ and every time $t$:

The letters stand for these quantities:

• $A$ is the amplitude, in m. It is the largest displacement.
• $\lambda$ is the wavelength, in m. It is the distance from one crest to the next.
• $k = \dfrac{2\pi}{\lambda}$ is the wave number, in rad/m.
• $f$ is the frequency, in Hz. It is the number of waves that pass a point each second.
• $\omega = 2\pi f$ is the angular frequency, in rad/s.
• $v = f\lambda = \dfrac{\omega}{k}$ is the wave speed, in m/s.

Positions are an array

numpy can work out a sine for a whole array of positions at once. np.linspace(0, 4, 9) gives nine positions from 0 m to 4 m, half a metre apart. The displacement of the whole rope is then one line:

It prints [0. 0.35 0.5 0.35 0. ]. Each number is the displacement of one point of the rope, in m. The rope rises to 0.50 m at 1.0 m, a quarter of a wavelength from the left end.

Putting time in

To make the wave move, subtract $\omega t$ inside the sine. A crest is the place where $kx - \omega t = \dfrac{\pi}{2}$. Solving for $x$ gives

The crest position grows steadily with time, at the rate $\omega / k$. That rate is the wave speed, $v = \omega / k = f\lambda$.

The minus sign makes the wave travel towards positive $x$. A plus sign, $kx + \omega t$, makes it travel the other way.

Finding a crest in code

np.argmax(y) gives the index of the largest number in y, and x[np.argmax(y)] is the position of that crest. If the array holds several equal crests, argmax gives the first one. A slice such as y[:200] keeps only the first part of the positions, so only one crest is left in it.

Practise with new numbers

Try it in Python

The program draws a wave at three times, finds the first crest each time, and works out the speed from how far the crest moved.

Common mistakes

• Leaving out the $2\pi$. Writing np.sin(x - f * t) is not a wave of frequency f. The sine needs k = 2 * np.pi / lam and w = 2 * np.pi * f.
• Using degrees. np.sin expects radians. np.sin(30) is not 0.5.
• Mixing up the sign. k * x - w * t travels towards positive $x$, and k * x + w * t towards negative $x$.
• Multiplying instead of dividing. The wave speed is w / k, not w * k.
• Forgetting that argmax finds one crest. Without a slice it returns only the first of several equal crests.

Remember these

• $y = A\sin(kx - \omega t)$ with $k = 2\pi/\lambda$ and $\omega = 2\pi f$.
• The wave speed is $v = f\lambda = \omega/k$.
• One line of numpy draws the whole rope: A * np.sin(k * x - w * t).
• x[np.argmax(y)] is where the first crest is.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why does the wave number have a $2\pi$ in it, $k = 2\pi/\lambda$?

2. A wave has a frequency of 5 Hz and a wavelength of 0.30 m. What is its speed?

3. Which way does $y = A\sin(kx + \omega t)$ travel?

4. What does x[np.argmax(y)] give?

5. np.sin(30) gives $-0.988$. Why is it not 0.5?