Before you read this, I suggest you read posts 18.8 and 18.10; it might also help to read post 18.11.
Let’s suppose a string or wire is fixed at A, is stretched tight by a force T (a tensile force, post 16.49) and is then clamped at B. This wire is shown in black. Now let’s displace the centre of the wire, downwards, so that the wire is bent into the lower red shape. If we then remove the displacing force the wire spontaneously moves upwards – it stored the work done in displacing it as potential energy; it then uses this potential energy to recoil – converting the kinetic energy into potential energy (see post 16.21 for more information on potential and kinetic energy). The wire behaves like a spring (post 16.49). However, just like the spring in post 18.11, it does not simply return to its original position – it oscillates. (Further information on why it oscillates is given in the appendix of post 18.11.) The wire moves from its lower red position, through its lower blue, lower green and black positions and then upwards, through its upper blue and upper green to its upper red position; it then moves downwards, in the reverse order. In the absence of damping (post 18.9), the oscillation will continue indefinitely.
The wire is then a simple harmonic oscillator vibrating with its natural frequency (post 18.8).
Notice also that the wire is moving up and down like a wave (post 18.10). But, unlike the water waves we met in post 18.10, it does not travel along the wire. The wave on the wire does not move along a direction of propagation – it is a standing wave. The ends A and B are fixed and so cannot move up and down – they are called nodes. The centre of the wire shows the maximum displacement – it is called an anti-node.
Let’s suppose the length AB of the wire is L. According to the picture, in post 18.10, defining wavelength, λ, only half a wave-length of the standing wave appears on the wire. So
λ = 2L.
Also according to post 18.10, the frequency of the wave is given by
f = v/λ = v/2L,
where v is its speed. Note that f is the natural frequency of the vibrating wire.
In the appendix, we see that the speed of the wave is
v = (T/σ)1/2,
where σ is the mass per unit length (measured in kg.m-1 in the SI system, post 16.13) of the wire. So σ is a bit like density, ρ, (post 16.44) except that it’s a mass divided by a length instead of a mass divided by a volume. (You can work out that, for a wire with a circular cross-section of radius r, that σ = πr2ρ.) If we substitute the expression for v into the expression for f, we obtain the result that the wire vibrates with a frequency
f = (1/2L)(T/σ)1/2.
So the longer the wire, the lower the natural frequency – the strings of a cello have a lower natural frequency than the strings of a violin. When you play a guitar or piano, you can decrease L, and so increase f, by clamping one end of the string under your finger. Increasing the tension of the wire increases f; this is what happens when you tighten or loosen the strings of a violin or a guitar when you are tuning it to vibrate with a target frequency. A piano tuner does the same thing when he or she tunes a piano. Finally, if the string or wire has a high value of σ, f will decrease; this is why all the strings on a guitar or a violin are not identical.
By now you will have guessed (if you didn’t know already) that the frequency of a vibrating string produces a sound with a required pitch – to make music. So why does the same musical note (a sound with a particular pitch or frequency) sound different when played on a violin, a guitar and a piano?
The vibrating wire is forced to have nodes at A and B because it is fixed at these positions. However, it can have antinodes in positions other than its centre. The pictures above show some more waves that can form on the wire when it is fixed at A and B. In the picture they are all shown with the same amplitude. But this won’t necessarily be the case. We expect the red wave to have the larger amplitude if we displace the wire nearer its centre. You can see that, for the blue wave, λ = L. The wavelengths, and therefore the frequencies, of the other waves can be related to L, as shown in the table below.
So, the general expressions for the allowed wavelengths and frequencies is
λn = (2/n)L aλn = (2/n)L and fn = (n/2L)(T/σ)1/2.
When n = 1, the string vibrates with its fundamental frequency; the other frequencies, that are related to it by having different values of the integer n, are its harmonics.
So, to return to our question, why do a violin, a guitar and a piano sound different when they’re playing the same note? In each case, the string is made to vibrate in a different way. In a violin, the string is displaced by a frictional force, when the bow rubs over it. In the guitar, the string is plucked and, in a piano, it is hit by a hammer. Each method of making the string vibrate leads to a different mix of harmonics and so to a different sound.
The performance of a real musical instrument is much more complicated than I’ve made it seem. In an electric guitar the vibration of the string is turned into an electrical signal that is amplified electronically. In an acoustic guitar, the sound is amplified by the vibration of the hollow body of the instrument and the air inside it. So, an electric guitar has an electrical amplifier and an acoustic guitar has an acoustic amplifier – the two different amplifiers affect the sound produced differently. As a result, an electric and an acoustic guitar sound different. Plucking a violin string (pizzicato) produces a different mix of harmonics, and a different sound, than when it is played with a bow. But a plucked violin sounds different to a plucked guitar, because they have very different acoustic amplifiers. A harpsichord sounds different to a piano, although it has a very similar acoustic amplifier, because its strings are plucked instead of being hit with a hammer.
In this post, we have seen how a vibrating wire (a simple harmonic oscillator) can be considered as a form of wave. This is an example of the link between simple harmonic oscillators and waves (post 18.11) and helps us to understand something about how stringed musical instruments work. However, we have yet to explore the nature of sound itself – the subject of my next post.
Related posts
18.11 Motion in a circle, the simple harmonic oscillator and waves
18.10 Waves
18.8 Natural frequency and resonance
18.7 The simple pendulum
18.6 The pendulum: a simple harmonic oscillator
17.13 Centripetal force
17.12 Going round in circles
16.14 Aliasing
Appendix
The purpose of this appendix is to show that v = (T/σ)1/2.
Let’s think about a short segment of the wire, of length ΔL, at the peak of the wave. If we make ΔL sufficiently small, it approximates to the arc of the circle, radius r, shown in the picture above.
The blue radius of the circle is perpendicular to the tangent (dashed line) at the mid-point of the segment. The green radii meet the ends of the segment and each makes an angle, θ, with the blue radius.
If we measure θ in radians
2θ = ΔL/r or θ = ΔL/2r
as explained in post 17.11.
The tension in the wire acts at both ends of the segment. It makes an angle of θ with the tangent. (You need to look at the picture to convince yourself of this, noting that the tangent is perpendicular to the blue radius). The horizontal components (post 16.50) of this tension act in opposite directions and cancel each other out. The vertical components are Tsinθ (post 16.50) and act at both sides of the wire. So the total vertical force, that acts towards the centre of the circle, is given by
2Tsinθ ≈ 2Tθ = TΔL/r.
The second step arises because the sine of a small angle is approximately equal to its value measured in radians; this is proved, by different methods, in appendix 1 of post 18.6 (noting that 1 – cos2θ = sin2θ) and in the appendix of post 18.7.
If we consider the segment to have a mass m and to be moving in a circular path of radius r, with speed v, this force provides the centripetal force (post 17.13) to maintain the motion, with the result that
TΔL/r = mv2/r = σΔLv2/r.
Multiplying the right and left-hand sides by r/ΔL gives the result that
v2 = T/σ.
Taking the square root of both sides of this equation gives the required result.
Thinking about Science with David Hukins 