Interactive Lecture Demonstration 18 – Standing Waves on a String

We will create a standing wave on a stretched string, by vibrating the string up and down with a mechanical oscillator. This will be a transverse wave, since the motion of the string will be up and down, whereas the wave crests will travel left to right (or vice versa) on the string.

As we will learn tomorrow, a standing wave actually consists of two different waves on the string. The two component waves travel in opposite directions but are otherwise identical (same speed, frequency, wavelength, amplitude). The wave traveling in one direction in this demonstration is the wave created by the mechanical oscillator at one end of the string. The wave traveling in the other direction is made when the first wave reflects from the other end of the string.

The wave that you will actually see on the string is the superposition (total) of the two component waves. We will see tomorrow that the superposition of the two traveling waves is a wave with crests that do not move left or right; that's why we call it a "standing wave."

We will use the following basic laws of physics relating to waves:

• For any wave traveling through a medium, the speed of the wave is determined solely by the ratio of two properties of the medium. One property describes how difficult to stretch (or compress) the medium is; the other property describes how much inertia the medium has. For a wave traveling on a string, those two properties are the tension T and the linear density M/L of the string:

v = √ T / (M/L).

• Although the wave crests move along the string with a speed that depends on the tension and linear density of the string, the individual particles of the string do not move along with the wave crests. Instead, the individual particles of the string vibrate up and down in simple harmonic motion. Thus, the vertical position vs time of any particle of the string is described by: y(t) = A cos(2 p f t), where f is the frequency of the wave. The amplitude A, as you will see, is different for different particles in a standing wave. We have assumed here that the equilibrium position of all the particles of the string is y = 0.

• For any periodic wave, the speed of the wave (i.e. the speed with which wave crests move along the medium) is related to the frequency and wavelength of the wave by: v = f l.

1. The diagram on the previous page represents a "snapshot" of the standing wave at one instant of time. Let's assume the snapshot occurs at an instant when the standing wave is at its maximum vertical displacement. What two properties of the wave can you see on this kind of diagram ? Label them on the diagram with appropriate variables.

2. Now observe the actual standing wave on the string. Measure and record the wavelength and amplitude of the wave. Also record the frequency of the vibration made by the oscillator.

3. Even though you can't see them, there are wave crests moving to the left and right on the string. Use your measurements to calculate what the speed of wave crests on the string must be.

4. The tension in the string is created by the mass hanging from the end of the string. Measure how much hanging mass there is, then do a net force problem to calculate what the tension in the string must be.

5. Since we now know the speed of the waves on the string and the tension in the string, we can calculate what the string's linear density must be. Do that now.

6. Now we will change the frequency of the oscillator. Predict what will happen to the wavelength of the standing wave if the frequency of oscillation is increased.

The speed of the waves on the string does not change if we only change the frequency of the oscillator. The speed only depends on the tension and linear density of the string, which we did not change. Therefore, if we increase the frequency of the wave, its wavelength must decrease in order for the wave's speed to remain constant.

When we increase the frequency, you should notice that the wave pattern looks very complicated until we reach a certain frequency when the wave pattern should again look sinusoidal. You are seeing the effect of interference between the waves traveling in opposite directions. For most frequencies, the two component waves add together to make a complicated superposition. Only for certain frequencies do they create a simple, sinusoidal standing wave pattern.

Here's how to predict what frequencies will make a standing wave. Notice that the ends of the string do not move up and down – they are more or less fixed in place due to being tied to something. As we will learn tomorrow, a standing wave will only be made in this situation if a complete number of half-wavelengths of the standing wave fit exactly onto the string. We can write the following equation that expresses that condition:

L = nl/2 where n = 1,2,3,... any positive integer

(condition for standing waves on a string fixed at both ends)

The length of the string must equal an integer number of half-wavelengths, so that the standing wave always has zero amplitude at the ends of the string.

The first standing wave we saw corresponded to the case n = 1. The wave we have just made corresponds to the case n = 2. You should see one complete wavelength of the wave between the ends of the string. Sketch below the standing wave pattern (i.e. a "snapshot" of the standing wave) for the n=2 case, and label on your sketch the wavelength of the wave.

7. How should the frequency of the new standing wave compare to the frequency of the first one ? Why ?

8. If we kept the string the same length, but increased the tension by adding more hanging mass, what would happen to:

a)      the wavelength of the standing wave pattern ? Why ?

b)      the speed of the wave crests ? Why ?

c)      the frequency of the standing wave pattern ? Why ?

9. How does the speed of each particle of the string compare with the speed of the wave crests ? To find out, let's now calculate the maximum speed of one particle of the string. Remember that the string moves up and down even though the wave moves along the string. We can think of each particle of the string as being a small mass m attached to a small spring with spring constant k. When we stretch a particle of the string in the transverse direction, the little spring will tend to pull the particle back to its equilibrium position. Using this analogy, we can find the maximum particle speed by applying the law of conservation of energy to one particle of the string. We should include the particle's kinetic energy and the elastic energy of the small spring. We will ignore gravitational energy here, since that would only complicate our calculation. As we saw in the Followup problems to ILD19, the maximum speed for a mass on a spring is the same whether the spring is vertical (when gravitational energy does change during the motion) or horizontal (when gravitational energy is unimportant since it doesn't change), so we are safe in ignoring it here.

Starting from conservation of energy, calculate now the maximum speed of a particle of the string. Compare your answer with the speed of the wave crests.