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11

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


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

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 ...


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 ...


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

$x_{max}$ 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 ...


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

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 ...


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

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 ...


2

Keith Thompson's answer, while good, makes it seem that there is some intrinsic difference between the black keys and white keys, and that only the white keys approximate integer harmonies. This is not really true. The white keys are just the C major scale, and the most important purpose the black-key/white-key distinction serves is to make it easy to see ...


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

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


2

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 ...


1

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 ...


1

There's an error in that the type of pipe for each of the two fundamental frequencies as described in your comment don't match the problem description. The pipe with a fundamental frequency of 440Hz is open-closed, and the pipe with a fundamental frequency of 660Hz is open-open. You actually said "closed-closed", which isn't even an option, but even if ...


1

The classical string equation that you are referring to, is formulated by making a number of assumptions, which include that the vibration of the string does not affect its tension. This makes Young's modulus irrelevant for results calculated from the idealized equation. In the real world, materials with low moduli of elasticity will follow the ideal ...


1

You can send sound waves of any wavelength into your instrument, but the trouble is that only specific wavelengths correspond to standing waves. The other waves will just die out because of destructive interference. When your sound wave collides with the closed end of a pipe for instance, it gets reflected back. If the wavelength is not right, the ...


1

There is no easy answer for this. It happens to be a good arrangement for many reasons. Several factors would have contributed to the original design. Western music is based on the diatonic scale. C D E F G A B is the diatonic scale on C. Musical staves, created by Gregorian monks to record their plainchants before the existence of keyboards, and still ...


1

A couple things play in here. First, the string is "closed" at both ends, meaning the ends are locked down and can't move. This means any resonant wavelength must have "nodes" , which is a contraction of "no displacement," at the ends. By comparison, the acoustic waves in an open-ended tube may have a node at one end, but the other end is unrestrained ...


1

Modern keyboards are tuned with each of the 12 half-step intervals between one octave and the next being equal, so a half-step is a ratio of the 12th root of 2 (about 1.059), and a whole-step is a ratio of the 6th root of 2 (about 1.122). But the ratios that "sound nice" musically tend to be ratios of small integers. 2:1 is an octave, of course. 3:2 is a ...


1

I'm not qualified to answer this question in detail...but I'd like to point out some things I've learned lately that may be helpful. Hilbert space is useful when you need an infinite dimensional space to characterize what you are studying and where each mode is orthogonal to the others. Great for Quantum Mechanics... Where I work, images, that is 2D ...



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