A sitar has a second set of thin strings, about a dozen, running under the main strings. Nobody plucks them. The player tunes each one to a note of the raga. When a main string plays one of those notes, the matching thin string starts to sound by itself. A thin string of any other note stays silent.
A wave on a string of fixed length
When a wave is confined to a finite length, it can no longer have just any frequency. Think of a guitar string held between the bridge and a fret. The two boundaries impose strict constraints. They force the wave to interact with its own reflections in a very specific way.
When the driving frequency is just right, the reflections interfere constructively, and the amplitude of the wave grows dramatically. Energy is then pumped into the string efficiently, and a stable standing wave of large amplitude forms. This powerful effect is called resonance. The frequencies that cause it are the string's resonant frequencies. The patterns of vibration that result are the string's normal modes.
A string of fixed length is in resonance when it is driven at one of its resonant frequencies. Its reflections then interfere constructively, the string absorbs energy efficiently, and a large, stable standing wave forms. Each such pattern is a normal mode of the string.
The physical mechanism of resonance
Why do only certain frequencies work? Let us follow the process in detail. Imagine that a source sends waves continuously into one end of a string fixed at both ends. The source may be a vibrating tuning fork or a hand. The string has length \(L\).
- The journey. A wave travels from the source to the far fixed end, a distance \(L\).
- The first reflection. The wave reflects from the fixed end. The reflection inverts it, as the chapter Superposition and Reflection showed.
- The return. The inverted wave travels back towards the source.
- The second reflection. The source end is also effectively fixed, so the wave reflects again and is inverted a second time. The twice-reflected wave is upright again.
- The crucial interference. The twice-reflected wave has now travelled a round trip of \(2L\). It moves in the same direction as the new waves that the source is making.
The critical question is this. Will the returning wave interfere constructively or destructively with the new wave being made?
- If the frequency is not resonant, the returning wave is out of phase with the new wave. Their interference is haphazard: sometimes constructive, sometimes destructive. No large amplitude builds up, and the string absorbs very little energy.
- If the frequency is resonant, the returning wave arrives back at the source perfectly in phase with the new wave. This perfect constructive interference raises the amplitude. The larger wave then makes another round trip and returns in phase again, to add to an even newer wave.
The amplitude does not grow for ever. It builds up until the energy pumped in by the source is balanced by the energy lost, for example to air resistance. A larger swing loses more energy in each cycle. So the growth slows, and the amplitude settles at the level where the gain and the loss are equal.
The source drives the string at a resonant frequency, so every returning wave arrives in phase. What happens to the amplitude over many round trips?
In the figure above the string vibrates in its simplest pattern, a single loop. One round trip takes \(2L/v\), and in that time the loop makes exactly one oscillation. After one round trip the amplitude is \(49\,\%\) of the dashed limit. After three it is \(86\,\%\), and after six it is \(98\,\%\). Each round trip adds less than the one before.
The condition for resonance
The two reflections invert the wave twice, and two inversions cancel. So the returning wave differs from the new wave only through the extra distance it has travelled, \(2L\). The two are in phase when that distance holds a whole number of wavelengths. This condition for perfect constructive interference is equation (12):
\[ 2L = n\lambda, \qquad n = 1, 2, 3, \dots \]
This is the fundamental condition for resonance on a string fixed at both ends.
A wave leaves the source end of a string fixed at both ends. It reflects at the far end and again at the source end. Compared with the wave that left, the returning wave is
A string \(1.2\,\text{m}\) long, fixed at both ends, resonates with \(n = 3\) in equation (12). Find the wavelength, in m.
A string \(0.60\,\text{m}\) long is fixed at both ends. Which wavelengths can resonate on it?
A string is driven at a frequency that is not one of its resonant frequencies. Why does it vibrate only weakly?
Boundary conditions: a string fixed at both ends
We can also analyse the final, stable standing waves that resonance produces. These patterns must obey the physical rules that the system imposes, called boundary conditions. For this string the rule is simple. The ends at \(x = 0\) and \(x = L\) are fixed and cannot move. Therefore both ends must be nodes.
A standing wave of equation (11) can exist on this string only if its shape function, \(2A\sin(kx)\), is zero at both ends. The derivation below applies the two conditions in turn.
\[ 2A\sin(k \cdot 0) = 0 \]
Why. the condition at \(x = 0\). It holds for every \(k\), since \(\sin 0 = 0\). Equation (11) already has a node at \(x = 0\), as Lesson 1 showed.
\[ \sin(kL) = 0 \]
Why. the condition at \(x = L\): \(2A\sin(kL) = 0\), and \(A\) is not zero. This is the boundary condition.
\[ kL = n\pi, \qquad n = 1, 2, 3, \dots \]
Why. the sine is zero only at whole multiples of \(\pi\). The value \(n = 0\) would make \(k = 0\), a flat string with no wave.
\[ \dfrac{2\pi}{\lambda_n}\,L = n\pi \]
Why. substitute \(k = 2\pi/\lambda\). Each value of \(n\) allows its own wavelength, so write it \(\lambda_n\).
\[ L = n\left(\dfrac{\lambda_n}{2}\right) \]
Read it. solve step 4 for \(L\). This is equation (13).
Equation (13) says that the string must hold a whole number of half wavelengths, that is, of loops. Multiply it by 2 and it reads \(2L = n\lambda_n\). This is exactly the condition found from the physical resonance argument, equation (12). So the stable standing waves are precisely those that satisfy the condition for constructive interference. Two different arguments, one about round trips and one about ends, give the same answer.
The round-trip argument gave \(2L = n\lambda\), and the boundary conditions gave \(L = n(\lambda/2)\). Why do the two arguments agree?
They describe the same wave in two ways. A standing wave is a wave and its reflections, so its ends must stay at rest, which means a whole number of half wavelengths fits between them. The same fit makes the round trip 2L a whole number of wavelengths, so the returning wave is in phase with the new one. A pattern that builds up by resonance must also be a pattern that the ends allow.
Now return to the sympathetic strings of the sitar. Each thin string has fixed ends, so it has its own resonant frequencies. The sound of a main string reaches it through the bridge and the air. Most frequencies arrive out of phase with the thin string's reflections, and nothing builds up. When a note matches one of its resonant frequencies, every round trip adds in phase. The thin string then grows loud enough to hear.
In his Two New Sciences of 1638, Galileo described a curious effect. Strike one string of an instrument, and another string tuned to the same note begins to vibrate by itself. A string tuned to a different note stays still. Galileo compared it with a heavy pendulum. Small puffs of breath, each given at the right moment of the swing, can set it swinging widely. Puffs given at random cannot. This is the round-trip argument of this lesson: timing, not force, builds the vibration.
- Thinking that a stronger source makes any frequency resonate.Resonance is about timing. Off resonance the returning waves arrive out of phase, and a stronger push only adds to the haphazard interference.
- Counting only one inversion in the round trip.A string fixed at both ends reflects the wave twice, once at each end. Two inversions cancel, so only the path \(2L\) sets the phase.
- Writing \(L = n\lambda\) for a string fixed at both ends.A whole number of half wavelengths fits between two nodes: \(L = n(\lambda/2)\), or \(2L = n\lambda\).