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15.2Standing waves on strings

Take a string of length L, tighten it and keep both ends fixed by attaching one end to a clamp and the other end to an oscillator. (The point on the string that is attached to the oscillator does not move relative to the oscillator so we have a node there.) In this case the end conditions are node–node. The oscillator creates travelling waves that move towards the fixed end of the string. The waves reflect at the fixed end, and so at any one time there are two identical travelling waves on the string travelling in opposite directions. As we saw, this is the condition for a standing wave to form. However, for the standing wave to fit on the string the right frequency has to be used, as we will see (see worked example 15.1). Different standing wave patterns will be established on the string depending on the frequency used. Figure 15.2 shows four possibilities. In the top diagram we see one loop, with two nodes and one antinode. This is the standing wave with the longest wavelength (and the lowest frequency): it is called the first harmonic the longest wavelength (and the lowest frequency) at which a standing wave forms. The higher harmonics standing wave with a frequency that is an integral multiple of the first harmonic frequency have more loops, and the wavelength decreases and the frequency increases, as shown in the lower part of Figure 15.2. In the figure, the symbol n stands for the number of the harmonic so that, for example, n = 3 indicates the third harmonic.

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Figure 15.2: Standing waves on a string with both ends fixed. The first four harmonics are shown. (The oscillator is not shown.)

It is important to realise that, for each harmonic, there is a definite relationship between the wavelength and the length of the string.

Remember that the distance between two consecutive nodes is half a wavelength. So for the first harmonic we have:

λ 1 2=L⇒λ 1=2 L=2 L 1

For the second harmonic:

2×λ 2 2=L⇒λ 2=L=2 L 2

For the third harmonic:

3×λ 3 2=L⇒λ 3=2 L 3

In general, we find that the wavelengths satisfy:

λ n=2 L n,n=1,2,3,4…

Figure 15.2 also gives the frequencies of the harmonics using the equation f=v λ, where v is the speed of the travelling waves making up the standing wave and λ the wavelength of the harmonic. The first harmonic has the lowest frequency. Notice the important fact that all harmonics have frequencies that are integral multiples of the first harmonic frequency.

Worked Example 15.1

The speed for waves on a string of length 1.20 m is 140 m s−1.

At what frequency should the oscillator be set so that the string vibrates in the first harmonic?

Answer

The wavelength in the first harmonic is twice the length of the string (1.20 × 2 = 2.40 m).

The required frequency is, therefore:

f=v λ=140 2.40≈58 Hz

Figure 15.3 shows successive positions of the string for the first harmonic. This diagram shows that different points on the standing wave oscillate with different amplitudes. The figure shows the shape of the string at different times (seven times in this case). There is an antinode midway between the two nodes.

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Figure 15.3: Different points on the string have different amplitudes.

CHECK YOURSELF 2

In Figure 15.3, which point on the string has the largest average speed in the course of one period?

Figure 15.4 shows how the string oscillates in its second harmonic. Here we have three nodes and two antinodes. The arrows represent the velocity vectors of points on the string. Notice that the direction of the velocity is the same for points within the same loop. We say that points within the same loop oscillate in phase. In the next loop the velocities are also in the same direction but opposite to those in the previous loop. We say there is a 180° phase difference between points in one loop and points in the next loop.

Figure 15.4: Points within consecutive nodes have velocity in the same direction. Points in the next loop have opposite velocity directions.

Figure 15.5 shows an actual string vibrating in the second harmonic.

Figure 15.5: This string vibrates in the second harmonic.

CHECK YOURSELF 3

In Figure 15.5 the tension in the string is increased, doubling the speed of the wave. The frequency of the oscillator and the length of the string are unchanged. What is the shape of the new standing wave?

Worked Example 15.2

A standing wave is set up on a string with both ends fixed. The frequency of the first harmonic is 150 Hz. The speed of the wave on the string is 240 m s−1 and the speed of sound in air is 340 m s−1.

Calculate:

athe length of the string

bthe wavelength of the sound produced.

Answer

aThe wavelength is given by:

λ 1=240 150=1.6 m

The wavelength of the first harmonic is 2L and so L = 0.80 m.

bThe sound will have the same frequency as that of the standing wave, i.e. 150 Hz. The wavelength of the sound is thus:

λ=340 150≈2.3 m

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TEST YOUR UNDERSTANDING

5 aDescribe how you would arrange for a string that is kept under tension, with both ends fixed, to vibrate in its second harmonic mode.

bDraw the shape of the string when it is vibrating in its second harmonic mode.

6A string is held under tension, with both ends fixed, and has a first harmonic frequency of 250 Hz. The speed of a transverse wave on a sting is given by V=T μ where T is the tension.

aBy finding the units of the quantity µ, identify µ.

bThe tension in the string is doubled. Predict the new frequency of the first harmonic.

7Strings X and Y have both ends fixed. The speed of transverse waves in X is double that in Y, and the length of X is half that of Y. Determine the ratio of the frequency of the first harmonic in X to that in Y.

8The wave velocity of a transverse wave on a string of length 0.500 m is 225 m s−1.

aDetermine the frequency of the first harmonic of a standing wave on this string when both ends are kept fixed.

bCalculate the wavelength of the sound produced in air by the oscillating string in part a. (Take the speed of sound in air to be 340 m s−1.)

9The velocity of a transverse wave on a string is 122 m s−1. The string has both ends fixed, and a standing wave of frequency 512 Hz is established on the string. What is the distance between consecutive nodes?

10A string has both ends fixed. A particular harmonic has frequency 340 Hz. The next harmonic has frequency 425 Hz. What is the frequency of the first harmonic?

11A string has both ends fixed and oscillates in the first harmonic. The amplitude in the middle of the string is 5.0 mm and the frequency of the harmonic is 480 Hz. What is the average speed of a point in the middle of the string during one full oscillation?