# Tag Info

46

The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular. If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The ...

46

Why not calculate it? Consider a string of length $L$, with its ends fixed at $x=\pm\frac{L}{2}$. Let's assume for convenience that at time $t=0$ the string is "plucked" at $x = 0$, so that the string displacement relative to its equilibrium position is given by $$f(x)=A\left|1-\frac{2x}{L}\right|.$$ The standing wave solutions to the wave equation ...

25

The answer is actually very dependent on how you pluck the string. If you pluck it closer to the center, you put more energy into the lower modes. Pluck it near either end, and you have more higher harmonics. And then there's the overtone techniques, which intentionally squelch lower harmonics, leaving only higher harmonics.

23

When you pluck the string you excite many many overtones, not just the fundamental. You can observe this by suppressing the fundamental. Pluck the string while holding a finger lightly at the center of the string. That point is an antinode for the fundamental and all odd harmonics, but a node for the even harmonics. Putting your finger at that point ...

20

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern ...

17

It's simple energy conservation. With an increase in harmonics, the frequency of vibration of the string increases. We know that each particle in the string is executing a simple harmonic motion with energy: $e=\frac{1}{2}m{\omega}^2A^2$ We have a continuous distribution of such oscillating masses, each oscillating with different amplitudes. Integrating them ...

16

Usually, the (sinusoidal) driven harmonic oscillator is damped, and the first two parts of your solution (which depend on the initial conditions, while the third term does not) are transient, i.e. not relevant after a short time. That the solution $$x(t) = \frac{F_0\sin(\omega t)}{m(\omega_0^2 - \omega^2)}$$ cannot be the "full" solution to the equation of ...

16

$y(t)=A \sin(\omega t+d)=A \cos(\omega t+d-\left(\frac\pi2\right))$ The two forms are the same except for the phase term, so which one you use depends on when you started your stopwatch. If you started off at max displacement at t=0, then you use the cosine form, if you start out with max velocity at t=0, then you use the sine form. If you are deducing ...

15

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like: $$A(t,x) = \sum_{i=0}^\infty A_i \sin(n\omega_i t - k_i x)$$ i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with ...

13

When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey $$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n},$$ where $\lambda_0$ is twice the ...

12

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^2}... 10 When you pluck a string or hit a drum or sound a not on a flute, the instrument and the air in and around it vibrate and this vibration propagates as sound waves in the air to your hear drum. When you hear an instrument being played, what you recognise as the note is the base frequency. 'C' corresponds to 261.6 Hz and is the same for a piano or a guitar. ... 10 Eigenmodes of a string have sinusoidal spatial form f_m(x) = C_m \sin(\pi m x/L), where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location x_0 means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, f(x) = A x/x_0 for x \le x_0 and f(x) = A (L-x)/... 10 A simple answer: the total energy of the vibration has to be finite. Given that we have an infinite number of possible modes of vibration (not only harmonics, but let's start with them), you need some distribution of the energy between few of them (in order to hear something at all) and you get less and less energy left for higher ones. p.s. you don't always ... 9 There are two ways to describe a sound wave. One is in terms of displacement of the medium and the other is in terms of pressure. This simple diagram shows that tthe two descriptions are 90^\circ out of phase with one another. Note that at a compression C where the pressure is a maximum the displacement of the particle is zero and the same is true at ... 9 Your mistake is in your initial assumption: When one pulls a string, it starts to oscillate and forms a standing wave with frequency$$f=f_0$$The only way this is true is if you could start the string with length L and wave speed v off with the exact shape of the sine wave of your fundamental frequency:$$y(x,t=0)=A\sin\left(\frac{\pi}{L}x\right)=A\...

9

For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency. For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}\cos (2\pi {{f}_{1}}... 8 Isn’t frequency how many cycles are completed per second, and isn’t the fundamental frequency only half a cycle When a string, fixed a both ends, vibrates in the fundamental mode, the perpendicular displacement$\phi_1(x,t)$of a point located at$x$along the length of the string is given by $$\phi_1(x,t) = A_1(t)\phi_1(x) = A_1\cos(2\pi f_1t + \varphi)\... 6 When you pluck a string it does not start out like the fundamental above. The string is pulled into a bent shape of two straight lines and an angle and it may not be bent at the middle. Releasing the bent string causes a bunch of harmonics of various amplitudes depending on how far off-center it was bent. (It can not return to the bent angle shape and the ... 6 As is often the case in physics, when the properties of string vibration are described invariably the string is treated as an idealized string. Among these idealizations: the string is treated as infinitely bendable. For the lower harmonics the error introduced by that simplification is acceptably small. That simplication fails for higher harmonics. On, say, ... 5 How does this happen? What's the physics behind it? If you fret the string and pluck it, the string will vibrate strongest at the middle point (between the bridge and the fret). This is the fundamental vibrational mode of the string. By placing your finger lightly at, e.g., the middle of the string and plucking (while quickly removing your finger), you ... 5 When you pluck a guitar string the potential you apply to the string is approximately a Dirac delta function. That is to say, the release of the string is a near instantaneous kick. One of the beautiful properties of the delta function is that its Fourier transform is unity. This means that it is made up of equal components of all frequencies. So, when ... 5 The response can be derived mathematically. Let u(x,t) denote the displacement of a point along the string at x at time t. The function obeys the wave equation in flat d=2 Minkowski space,$$\frac{\partial^2 u(x,t)}{\partial t^2} - v^2 \frac{\partial^2 u(x,t)}{\partial x^2}=0$$If we pinch the string at the middle, this corresponds to a condition ... 5 Yes, it most certainly can. It's much easier to visualize if you consider a length of flexible steel or plastic. You can shake it a bit, then toss it in the air so it's not constrained, and it will (if properly initiated) vibrate at a resonant wavelength. I think the confusion most people will get from your question is that everyday string is "floppy," i.... 5 You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ... 5 I can't be certain that this is generating your peaks, but any tone that starts and stops won't be 100% pure; a pure tone has no beginning and no end. Consider a tone that starts at 0 at t=0, vibrates for time \tau, and then turns off. As an equation, that looks like this:$$y(t) = \sin(2\pi f t)\, \Theta\left(\tau-t\right)\, \Theta(t),$$where \Theta(... 5 Your sampling rate is 48k at 55Hz, so each period is 872.73 samples. The size of your FFT is 65536. It fits 75.093 period of the signals. The algorithm takes 75 periods to plot the chart. This leaves 0.093 periods between consequitive FFT transforms. 0.093 periods at 55Hz correaponds to a frequency of 5.1Hz that matches the ghost frequency that you see ... 5 A standing wave on a string can be thought of as a traveling wave that is bouncing back & forth in one dimension. Every time it reaches one of the ends, it is reflected, either inverted or upright depending on what conditions you have at the end. Assuming you have the same boundary conditions at both ends, this means that by the time the traveling wave ... 5 You start with a triangular form, which has its fourier series. Let's say the initial shape is f(x):$$f(x)=\sum_n a_n \sin \frac{\pi n x}{L}$$where$n$counts the modes ($1$is fundamental). So initial shape determines the harmonic content$a_n\$ of certain modes. If you pluck in the middle, you will put more of the fundamental into the initial spectrum, ...

5

If there are overtones, then the "wave" is not just of the main frequency, but is more complicated. In the linear approximation the tone and overtones may come together - in a superposition, due to linearity of the wave equation (which admits very complicated "wave" profiles). You may excite any overtone without exciting the fundamental tone, so it is not ...

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