Fourier Series and Mode Decomposition
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You should be comfortable with the concept of normal modes for a string fixed at both ends and understand that the general solution to the wave equation is a superposition of these modes. Familiarity with basic trigonometric identities and definite integration is assumed.
Why this matters
Fourier decomposition allows us to break down any complex wave shape into simple, independent sine waves.
In the previous lecture, we established that a string fixed at both ends can vibrate in specific patterns called normal modes. However, real-world disturbances—such as plucking a guitar string or striking a bell—rarely match a single pure sine wave. Instead, they create complex, arbitrary shapes. The power of the wave equation lies in its linearity: if two shapes are valid solutions, their sum is also a valid solution. This principle allows us to treat any initial displacement as a sum of these fundamental modes. By decomposing the motion into its constituent frequencies, we can analyze energy distribution, predict how different parts of the wave decay at different rates due to damping, and understand the spectral content of the sound produced. This mathematical tool is not merely a trick for strings; it is the foundation for understanding light, quantum states, and signal processing.
The Idea of Superposition
Imagine a string stretched between two points. If you pluck it in the middle, it forms a triangle. This triangular shape is not one of the natural "pure" vibrations of the string. However, we can think of this triangle as being built by adding up several simpler waves. The first wave is a single arch (the fundamental mode). The second is a wave with a node in the middle (the second harmonic). The third has two nodes, and so on. If we add these waves together with the right strengths, they perfectly reconstruct the original triangle. Because the wave equation is linear, these added waves do not interfere with each other's evolution; they simply pass through one another. This means we can solve the complex problem of the triangle by solving many simple problems of sine waves and then adding the answers. This is the essence of Fourier analysis: any complex shape can be viewed as a stack of simple, independent oscillations.
Orthogonality of Modes
To determine exactly how much of each mode is present in our initial shape, we rely on a mathematical property called orthogonality. For a string of length \(L\) fixed at \(x=0\) and \(x=L\), the normal modes are given by \(\phi_n(x) = \sin\!\left(\frac{n\pi x}{L}\right)\). These functions are orthogonal, meaning that if you multiply two different modes together and integrate over the length of the string, the result is zero. This is analogous to how the dot product of two perpendicular vectors is zero. This property allows us to "isolate" a specific mode from the sum. If we have a general displacement \(y(x)\) written as a sum of modes, multiplying by a specific mode \(\sin\!\left(\frac{m\pi x}{L}\right)\) and integrating will cancel out all terms except the one corresponding to \(m\). This makes it possible to calculate the coefficient for each mode independently.
Calculating Modal Amplitudes
Using the orthogonality condition, we can derive a formula for the amplitude \(b_n\) of the \(n\)-th mode. We start with the expansion \(y(x) = \sum_{k=1}^{\infty} b_k \sin\!\left(\frac{k\pi x}{L}\right)\). To find a specific \(b_n\), we multiply both sides by \(\sin\!\left(\frac{n\pi x}{L}\right)\) and integrate from \(0\) to \(L\). The left side becomes \(\int_0^L y(x) \sin\!\left(\frac{n\pi x}{L}\right) dx\). On the right side, due to orthogonality, every term in the sum vanishes except the one where \(k=n\). That surviving term is \(b_n \int_0^L \sin^2\!\left(\frac{n\pi x}{L}\right) dx\), which evaluates to \(b_n \frac{L}{2}\). Solving for \(b_n\) gives us the projection formula. This integral is essentially a weighted average of the initial shape, where the weight is the shape of the mode itself. Modes that look more like the initial shape will have larger coefficients.
Time Evolution of the Modes
Once we have the spatial amplitudes \(b_n\), we can reconstruct the full motion \(y(x,t)\). Each mode oscillates in time at its own natural frequency \(\omega_n = \frac{n\pi v}{L}\), where \(v\) is the wave speed. The general solution is a sum of these time-dependent terms. If the string is released from rest, the initial velocity is zero, which means the time dependence must be a cosine function (since the derivative of cosine is zero at \(t=0\)). The full solution is therefore \(y(x,t) = \sum_{n=1}^{\infty} b_n \sin\!\left(\frac{n\pi x}{L}\right) \cos\!\left(\frac{n\pi v t}{L}\right)\). This equation tells us that the complex motion of the string is simply the sum of independent harmonic oscillators. The fundamental mode (\(n=1\)) oscillates slowest, while higher modes oscillate faster. This separation of space and time is a direct consequence of the linearity of the wave equation and the boundary conditions.
The derivation
Fourier Decomposition into Normal Modes
For a uniform string of length \(L\) fixed at both ends, the normal modes \(\phi_n(x)=\sin\!\left(\frac{n\pi x}{L}\right)\), \(n=1,2,3,\dots\), form a complete orthogonal set on \([0,L]\). Any admissible transverse configuration \(y(x)\) obeying the boundary conditions expands uniquely as \(y(x)=\sum_{n=1}^{\infty} b_n\,\sin\!\left(\frac{n\pi x}{L}\right)\), and each modal amplitude is recovered by projection, \(b_n=\frac{2}{L}\int_0^L y(x)\,\sin\!\left(\frac{n\pi x}{L}\right)dx\).
A common misconception is that every Fourier series shows the Gibbs overshoot. The Gibbs phenomenon (an overshoot of about 9% of the jump that never dies away as more terms are added) occurs only at a jump in the function, where the series converges to the average of the left and right limits. A plucked string has a corner but no jump, so its series converges uniformly and its coefficients fall off as \(1/n^2\); that is why low harmonics dominate the sound of a plucked string. C
Worked example
Problem. A uniform string of length \(L = 1.0\ \mathrm{m}\) is fixed at both ends. It is plucked at the midpoint, creating an initial triangular shape defined by \(y(x) = 2h x/L\) for \(0 \le x \le L/2\) and \(y(x) = 2h (L-x)/L\) for \(L/2 \le x \le L\), where \(h = 0.01\ \mathrm{m}\) is the maximum displacement. Calculate the amplitude \(b_1\) of the fundamental mode.
- Identify the symmetry. The function is symmetric about \(x = L/2\). The sine functions \(\sin\!\left(\frac{n\pi x}{L}\right)\) are also symmetric about \(L/2\) for odd \(n\) and antisymmetric for even \(n\). Since the initial shape is symmetric, all even coefficients \(b_n\) will be zero. We only need to calculate odd \(n\).
- Set up the integral for \(b_1\). Using the formula \(b_1 = \frac{2}{L} \int_0^L y(x) \sin\!\left(\frac{\pi x}{L}\right) dx\). Due to symmetry, we can integrate from \(0\) to \(L/2\) and multiply by 2.
- Substitute the expression for \(y(x)\) in the first half: \(y(x) = 2h x/L\). The integral becomes \(b_1 = \frac{2}{L} \cdot 2 \int_0^{L/2} \frac{2h x}{L} \sin\!\left(\frac{\pi x}{L}\right) dx = \frac{8h}{L^2} \int_0^{L/2} x \sin\!\left(\frac{\pi x}{L}\right) dx\).
- Perform integration by parts. Let \(u = x\) and \(dv = \sin\!\left(\frac{\pi x}{L}\right) dx\). Then \(du = dx\) and \(v = -\frac{L}{\pi} \cos\!\left(\frac{\pi x}{L}\right)\). The integral is \(\left[ -\frac{Lx}{\pi} \cos\!\left(\frac{\pi x}{L}\right) \right]_0^{L/2} - \int_0^{L/2} \left( -\frac{L}{\pi} \cos\!\left(\frac{\pi x}{L}\right) \right) dx\).
- Evaluate the boundary terms. At \(x = L/2\), \(\cos\!\left(\frac{\pi}{2}\right) = 0\). At \(x = 0\), the term is 0. So the boundary part is 0. The remaining integral is \(\frac{L}{\pi} \int_0^{L/2} \cos\!\left(\frac{\pi x}{L}\right) dx = \frac{L}{\pi} \left[ \frac{L}{\pi} \sin\!\left(\frac{\pi x}{L}\right) \right]_0^{L/2}\).
- Calculate the final value. \(\sin\!\left(\frac{\pi}{2}\right) = 1\) and \(\sin(0) = 0\). The integral value is \(\frac{L^2}{\pi^2}\). Substituting back into the expression for \(b_1\): \(b_1 = \frac{8h}{L^2} \cdot \frac{L^2}{\pi^2} = \frac{8h}{\pi^2}\).
- Insert numerical values. \(h = 0.01\ \mathrm{m}\), \(L = 1.0\ \mathrm{m}\). \(b_1 = \frac{8 \times 0.01}{\pi^2} \approx \frac{0.08}{9.87} \approx 0.0081\ \mathrm{m}\).
Units check. \(b_1 = 8h/\pi^2\) has the units of \(h\), metres, as an amplitude must. The integral \(\int x\sin(\pi x/L)\,dx\) has units \(\mathrm{m^2}\) and the prefactor \(8h/L^2\) has units \(\mathrm{m^{-1}}\), giving metres.
Three quick tests
- If a string is plucked at \(x = L/3\), will the second harmonic (\(n=2\)) be present?
Answer
Yes. The shape is not symmetric about the center in a way that cancels even modes, and the projection integral for \(n=2\) will be non-zero. - Why do higher modes decay faster in a real string?
Answer
Higher modes involve more curvature and higher frequencies, leading to greater energy loss per cycle due to internal friction and air resistance, which are often frequency-dependent. - What is the value of \(b_n\) if the initial shape is exactly \(\sin\!\left(\frac{3\pi x}{L}\right)\)?
Answer
\(b_3 = 1\) and all other \(b_n = 0\). The shape is already a single normal mode.
Problem set for PU-103 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.