Chapter 3, Problem 80RE

To determine

To calculate: y’(r), y’(l),y’(d), and y’(T).

Answer to Problem 80RE

  $y\text{'}(r)=-\frac{1}{2l{r}^2}\sqrt{\frac{T}{\pi d}}$ .

  $y\text{'}(l)=-\frac{1}{2r{l}^2}\sqrt{\frac{T}{\pi d}}$ .

  $y\text{'}(d)=-\frac{1}{4rl}\sqrt{\frac{T}{\pi d^3}}$ .

  $y\text{'}(T)=-\frac{1}{4rl}\sqrt{\frac{1}{\pi dT}}$ .

Explanation of Solution

Given information: The fundamental frequency of the Piano string or wire is,

  $y=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}$. Where,

  r : the radius of the wire in centimeters;

  l: the length in centimeter;

  d : the density of the wire in grams per cubic centimeter;

  T : the tension (force) holding the wire taut in dynes.

  If keep all the variable fixed except one, then y can be alternatively thought of as four different function of one variable, y(r ),y(l),y(d),y(T).

Formula used:

  $\frac{d(x^n)}{dx}=n{(x)}^{n-1}.$

Calculation:

  $\begin{array}{l}y=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}.\\ \frac{dy}{dr}=y\text{'}(r)=\frac{d(\frac{1}{2rl}\sqrt{\frac{T}{\pi d}})}{dr}.\\ y\text{'}(r)=\frac{1}{2l}\sqrt{\frac{T}{\pi d}}\frac{d(\frac{1}{r})}{dr}.\\ y\text{'}(r)=\frac{1}{2l}\sqrt{\frac{T}{\pi d}}(-\frac{1}{r^2}).\\ y\text{'}(r)=-\frac{1}{2l{r}^2}\sqrt{\frac{T}{\pi d}}\mathrm{................}(1)\end{array}$.

  $\begin{array}{l}y=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}.\\ \frac{dy}{dl}=y\text{'}(l)=\frac{d(\frac{1}{2rl}\sqrt{\frac{T}{\pi d}})}{dl}.\\ y\text{'}(l)=\frac{1}{2r}\sqrt{\frac{T}{\pi d}}\frac{d(\frac{1}{l})}{dl}.\\ y\text{'}(l)=\frac{1}{2r}\sqrt{\frac{T}{\pi d}}(-\frac{1}{l^2}).\\ y\text{'}(l)=-\frac{1}{2r{l}^2}\sqrt{\frac{T}{\pi d}}\mathrm{..........................}(2)\end{array}$

  $\begin{array}{l}y=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}.\\ \frac{dy}{dd}=y\text{'}(d)=\frac{d(\frac{1}{2rl}\sqrt{\frac{T}{\pi d}})}{d(d)}.\\ y\text{'}(d)=\frac{1}{2rl}\sqrt{\frac{T}{\pi }}\frac{d(\sqrt{\frac{1}{d}})}{d(d)}.\\ y\text{'}(d)=\frac{1}{2rl}\sqrt{\frac{T}{\pi }}(-\frac{1}{2d^{\frac{3}{2}}}).\\ y\text{'}(d)=-\frac{1}{4rl}\sqrt{\frac{T}{\pi d^3}}\mathrm{..........................}(3)\end{array}$

  $\begin{array}{l}y=\frac{1}{2rl}\sqrt{\frac{T}{\pi d}}.\\ \frac{dy}{dT}=y\text{'}(T)=\frac{d(\frac{1}{2rl}\sqrt{\frac{T}{\pi d}})}{d(T)}.\\ y\text{'}(T)=\frac{1}{2rl}\sqrt{\frac{1}{\pi d}}\frac{d(\sqrt{T})}{d(T)}.\\ y\text{'}(T)=\frac{1}{2rl}\sqrt{\frac{1}{\pi d}}(\frac{1}{2}{T}^{-\frac{1}{2}}).\\ y\text{'}(T)=-\frac{1}{4rl}\sqrt{\frac{1}{\pi dT}}\mathrm{..........................}(3)\end{array}$