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# The frequency of vibrations of a vibration violin string is given by$f = \frac {1}{2L} \sqrt {\frac {T}{p}}$where $L$ is the length of the string, $T$ is its tension, and $p$ is its linear density. [See Chapter 11 in D.E. Hall, Musical Acoustics, 3rd ed. (Pacific Grove, CA; Brooks/Cole, 2002).](a) Find the rate of change of the frequency with respect to (i) the length (when $T$ and $p$ are constant), (ii) the tension (when $L$ and $p$ are constant), and (iii) the linear density (when $L$ and $T$ are constant).(b) The pitch of a note (how high or low the note sounds) is determined by the frequency $f.$ (The higher the frequency, the higher the pitch,) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note. (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg, (iii) when the linear density is increased by switching to another string.

## A)$\frac{d f}{d L}=\frac{-1}{2 L^{2}} \sqrt{\frac{T}{\rho}} \quad \frac{d f}{d T}=\frac{1}{4 L \sqrt{T \rho}} \quad \frac{d f}{d \rho}=\frac{-\sqrt{T}}{4 L \rho^{3 / 2}}$$\\$B)increase, increase, decrease

Derivatives

Differentiation

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Missouri State University

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University of Michigan - Ann Arbor

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University of Nottingham

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Idaho State University

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### Video Transcript

here we have a frequency equation and in the first part of part A, we're going to find d f D L. The rate of change of the length and the value of tea and the value of row are going to be held constant. So I'm going to rewrite the equation with all the constants together, 1/2 times the square root of tea over row. That's all a constant. And then the length is to the negative first power. So we're taking the derivative of that and we would bring down the negative once we would have negative 1/2 square root tea over row and then l would be to the negative second power so we can rewrite that as a negative one over to l squared times a square root of tea over row. Okay, so that takes care of part one of part a. Now let's move on to part two of Part A, where we're going to find d f d. L. The rate of change of the frequency, the ftt, the rate of change of the frequency with respect to t, which is tension. So once again, I'm going to rewrite my frequency equation first, and I'm going to write it as one over to L Times. Well, let me go ahead and put the square root of row on the bottom of that constant. And then that's going to be times t to the 1/2. Since we have square root tea, it's teach of the 1/2. Okay, so we're going to find the derivative of that. So are constant is going to be multiplied by 1/2 with to bring down the 1/2 so have 1/2 times are constant, one over to l square root row, and then T is going to be raised to the negative 1/2 power. So when we simplify that, we end up with 1/4 l Square root row T, and that's the second of the third, the second of three parts of part. A third part is we want to find the rate of change of the frequency with respect to row and row stands for the linear density. Okay, so once more, let's rewrite the original function, and we have square root tea over to El. That is all of our constant in this case, and we have a row to the negative 1/2. Okay, so we're gonna bring down the negative 1/2. We have negative 1/2. Let's see if we could make that negative 1/2 look a little bit better. Something like that doesn't want to cooperate with me very well right now. But there we go. Okay. So I bring down the negative 1/2 and then we have that multiplied by our constant, and then we need to raise row to the negative three halfs power. And if we want to change that back into radicals, we have negative square root tea over to l Times Square Root of roe. Cute. That's probably good enough for simplifying. Okay, so now we're on to part B. I'll switch to a different color. So we confined things better, and it says, use the signs of each of the derivatives we just found to determine what happens when and then there are several situations. Okay. So noticed that from the first part of part A, um the length was changing, and we had a negative derivative. That means the frequency is decreasing. Length changes frequency decreases. Okay. And then from the second part, notice that we had a positive derivative. So that means that the frequency is increasing. So the tension has increased the frequency increases. And then for the third part when the linear density increased, noticed that we have a negative derivative again. And so the density changed, It changed and the frequency decreased.

Oregon State University

#### Topics

Derivatives

Differentiation

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp