47

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 ...


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 ...


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 ...


15

$y(t)=A \sin(\omega t+d)=A \cos(\omega t+d-\left(\frac\pi2\right))$ 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 ...


11

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 ...


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

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

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)/...


9

Let's look at frequency instead of notes. Let's say the string has a natural frequency of $100 Hz$ and that harmonics are present when you pluck it. Then, the frequency content of the sound will be of the form: $a_1 \cdot 100 Hz + a_2 \cdot 200 Hz + a_3 \cdot 300 Hz + ... $ Now, let's say you fret this string halfway such that the natural frequency ...


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

A guitar string produces harmonics because it vibrates in a non-linear fashion. An electronic oscillator can be made to generate a much purer form of vibration (near sinusoidal) than a mechanical device such as the guitar string. Hence its harmonic level, while not zero, is much lower. For example, the harmonic distortion of a guitar string is probably on ...


7

Frequency is just a way of analyzing a time dependent motion. Consider plucking a string by first pulling one point on the string away from its equilibrium. The string shape will be like a triangle, two straight bits of string coming away from where your finger is holding the string, but meeting at a slight angle where your finger holds the string. That ...


7

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 ...


7

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 ...


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

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 ...


5

It depends on how you pluck the string. Typically when we pluck a string the string immediately before release looks something like this (I've exaggerated the vertical scale for clarity): When you pluck the string you pull it into a triangular shape. I've drawn the fundamenal sine wave in the background as a red dashed line, and you can see the shape of a ...


4

The physiology of human ear (and perhaps brain) makes sounds with frequency ~3000 Hz sound louder than higher and lower frequencies, for same sound wave pressure perturbation; see https://en.wikipedia.org/wiki/Equal-loudness_contour


4

Clearly the motion of the mass can not be described by a single sine! What's going on here? The general solution to the simple harmonic oscillator is the sum of the unforced response (homogeneous solution) and the forced response. The homogeneous solution is $$x_h(t) = x_h(0) \cos(\omega_0 t) + \frac{\dot x_h(0)}{\omega_0}\sin(\omega_0 t)$$ Thus, for ...


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