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12

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

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 ...

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

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

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

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

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 speed of sound should apply to $v$ because the sound waves are travelling through the air after it leaves the organ pipe. The speed of sound is approximated by the following formula: $$v = 331.3 + 0.606T$$ Where $T$ is the temperature in degrees Celsius, and $v$ is the velocity in meters per second. In your case, suppose you're at room temperature ...

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