# Tag Info

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 ...

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 ...

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. ...

7

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 ...

6

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 ...

6

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, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^... 5 Frequency is just a way of analyzing a time dependent motion. Consider plucking a string by first pulling one point on the string away from its equilibrium. The string shape will be like a triangle, two straight bits of string coming away from where your finger is holding the string, but meeting at a slight angle where your finger holds the string. That ... 4 It seems that the harmonic (integer multiple) overtones of a sound usually all have the same phase. Is this true...? No, I don't think this is generally true, although it may be true for certain instruments. What led you to believe this? In trumpet tones, for example, the different harmonics come up at different times during the attack, so it seems ... 4 When you release the plucked string, its shape is momentarily triangular: tied down at the ends and pointed at the location of your finger. But the solutions to the wave equation are not triangle functions, but sinusoidal functions, whose displacements from rest obey$$y_n(x) \propto \sin \frac{2\pi x}{\lambda_0 / n},$$where \lambda_0 is twice the ... 4 Clearly the motion of the mass can not be described by a single sine! What's going on here? The general solution to the simple harmonic oscillator is the sum of the unforced response (homogeneous solution) and the forced response. The homogeneous solution is$$x_h(t) = x_h(0) \cos(\omega_0 t) + \frac{\dot x_h(0)}{\omega_0}\sin(\omega_0 t)$$Thus, for ... 4 You always make harmonics when you pluck a string. You can change the spectrum of the harmonics --- that is, their relative intensities --- by changing where on the guitar string you pluck. The usual place to strum a guitar is near the sound hole, about a third of the way from the bridge. If you strum near the center of the string (12th/octave fret) you ... 4 You're wondering why pressure nodes form at an open end of a tube. The answer is, they don't! It's just a reasonably good approximation. Physically, consider the air molecules at the center of the tube. Since they're far away from the edges, there's no way for them to "know" exactly when the tube ends, so the sound wave must "leak out" slightly. The ... 3 However interesting, your question is probably too broad. When it comes to perception (so not just simple objective values), this topic is actually not perfectly understood in general. Always remember, that perception of any parameter of the sound is nonlinear and dependent on other parameters (e.g. you need to consider a pitch of the tone, when you ... 3 When you pluck a string it does not start out like the fundamental above. The string is pulled into a bent shape of two straight lines and an angle and it may not be bent at the middle. Releasing the bent string causes a bunch of harmonics of various amplitudes depending on how far off-center it was bent. (It can not return to the bent angle shape and the ... 3 Open organ pipe is the one with two open ends, and instead of the formula you mention you need to use$$L=n\frac{v}{2f_n}$$where f_n is the frequency of the {n^{th}} mode, and n=1,2,3,... your formula is for a closed organ pipe (with one open and one closed end). EDIT Because the number of half-wavelengths (\lambda /2) need to be an integral ... 3 The physiology of human ear (and perhaps brain) makes sounds with frequency ~3000 Hz sound louder than higher and lower frequencies, for same sound wave pressure perturbation; see https://en.wikipedia.org/wiki/Equal-loudness_contour 3 x_{max} is the amplitude of the oscillations, and yes, {\omega}t - \varphi is the phase. We know that the period T, is the reciprocal of the frequency f, or$$T = 1/f$$We also know that \omega, the angular frequency, is equal to 2\pi times the frequency, or$$\omega = 2{\pi}f$$From here, we can use the initial conditions to find the ... 3 There is a technique called flageolet where you damp the string with a finger laid lightly onto the site at the node of a higher harmonic. You do not press the string to the fretboard but just damp the string at a position, where there is a node of the specific harmonic. When you now pluck the string all harmonics, which do not have a node at the specified ... 3 Pitch, in music, is equivalent to frequency. How often the wavefore cycles. This is usually defined by length, i.e. how long the string is, how long the pipe is, etc. It can also be affected by the tension (how tight the string is.) Timbre, the sound of a specific instrument, is defined by the "shape" of the wavefore, whether spikes, round, square, or ... 3 Yes, it most certainly can. It's much easier to visualize if you consider a length of flexible steel or plastic. You can shake it a bit, then toss it in the air so it's not constrained, and it will (if properly initiated) vibrate at a resonant wavelength. I think the confusion most people will get from your question is that everyday string is "floppy," i.... 3 A few observations. First - if you record sound for a short time, the bandwidth of the sample will result in a smearing of the peaks. This only really matters if the sample is very short - with a 1 second sample you would have 1 Hz resolution, but if you sample for 0.01 second, the bandwidth is 100 Hz. Second, you are using a scale that is quite compressed ... 3 There are two ways to describe a sound wave. One is in terms of displacement of the medium and the other is in terms of pressure. This simple diagram shows that tthe two descriptions are 90^\circ out of phase with one another. Note that at a compression C where the pressure is a maximum the displacement of the particle is zero and the same is true ... 3 The equations of motion for the position determine the accelerations: they are second-order differential equations in time:$$\vec F = m\vec a = m\ddot{\vec x }$$So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the F=ma formula ... 2 The easiest way to determine maximum and minumums of a function is to set the derivative equal to zero. Thus, in this case, setting the equation for acceleration equal to zero and solving for the variables of interest will give you what you want. Thus, in this case we have the equation for position:$$ x = a \sin(\omega t + \phi) $$One way to see the ... 2 Like Jarosław Komar commented, you are using the wrong value for n. It is also easy to visualize this by looking at what the longest standing wave would look like in an air column with only one open end: Where the wavelength, \lambda, is defined as:$$ \lambda=\frac{v}{f} $$So the fundamental frequency would require a n=0, since the length of the air ... 2 The response can be derived mathematically. Let u(x,t) denote the displacement of a point along the string at x at time t. The function obeys the wave equation in flat d=2 Minkowski space,$$\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 ...

2

Hopefully David Bar Moshe can give a more rigorous explanation in terms of cohomology, but I have the following intuitive understanding of the difference between the two situations. In the Aharonov-Bohm effect, the particle is constrained to move around an (effectively) infinite solenoid. It then suffices to consider a problem on a plane, but with a hole in ...

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