Chapter 16, Problem 84P

(a)

To determine

To Find:The maximum kinetic energy of the wire.

Explanation of Solution

Given:

Length of the wire, $l=2.00\text{m}$

Tension in the wire, $T=40.0\text{N}$

Mass of the wire, $m=0.100\text{kg}$

At the midpoint, amplitude is $A=2.00\text{cm = 0}\text{.02 m}$

Formula Used:

Maximum kinetic energy of the wire can be obtained by:

  $K.E_{\max }=\frac{1}{4}m{\omega }^2{A}^2$

Here, m is the mass, $\omega$ is the angular frequency and A is the amplitude of the wave.

  $\omega =2\pi f$

Here, f is the frequency which can be obtained by:

  $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

  $\mu =\frac{m}{l}$

Calculations:

Find the mass per unit length:

  $\begin{array}{c}\mu =\frac{m}{l}\\ =\frac{0.100\text{kg}}{2.00\text{m}}\\ =0.05\text{kg/m}\end{array}$

Now calculate the frequency of the vibrating wire in fundamental mode:

  $\begin{array}{c}f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}\\ =\frac{1}{2\times 2.00}\sqrt{\frac{40.0}{0.1/2.00}}\\ =7\text{Hz}\end{array}$

The angular frequency is:

  $\begin{array}{c}\omega =2\pi f\\ =2\pi \left(7.00\right)\\ \approx 44\text{rad/s}\end{array}$

Now substitute all the known values to find the maximum kinetic energy of the wire:

  $\begin{array}{c}K.E_{\max }=\frac{1}{4}m{\omega }^2{A}^2\\ =\frac{1}{4}\left(0.100\right){\left(\text{44}\text{rad/s}\right)}^2{\left(0.02\text{m}\right)}^2\\ =0.0194\text{J}\\ \text{=19}\text{.4}\text{mJ}\end{array}$

Conclusion:

Thus, the maximum kinetic energy of the wire is $\text{19}\text{.4}\text{mJ}$ .

(b)

To determine

To Find: The kinetic energy of the wire at the instant when transverse displacement is given by $y=0.0200\sin \left(\frac{\pi }{2}x\right)$ .

Explanation of Solution

Given:

Length of the wire, $l=2.00\text{m}$

Tension in the wire, $T=40.0\text{N}$

Mass of the wire, $m=0.100\text{kg}$

At the midpoint, amplitude is $A=2.00\text{cm = 0}\text{.02 m}$

Displacement, $y=0.0200\sin \left(\frac{\pi }{2}x\right)$

  $0.00\text{m}\le x\le 2.00\text{m}$

Formula Used:

Wave equation of standing wave in fundamental mode:

  $y=A\sin \left(kx\right)\cos \left(\omega t\right)$

Calculations:

Compare the given displacement and the wave equation:

  $y=0.0200\sin \left(\frac{\pi }{2}x\right)$ and $y=A\sin \left(kx\right)\cos \left(\omega t\right)$

  $\begin{array}{l}\Rightarrow \cos \left(\omega t\right)=1\\ \Rightarrow \omega =0\\ \therefore K.E=0\end{array}$

Conclusion:

Thus, the kinetic energy at the given instant would be zero.

(c)

To determine

To Find: The value of x for which the average value of the kinetic energy per unit length is the greatest.

Explanation of Solution

Given:

Length of the wire, $l=2.00\text{m}$

Tension in the wire, $T=40.0\text{N}$

Mass of the wire, $m=0.100\text{kg}$

At the midpoint, amplitude is $A=2.00\text{cm = 0}\text{.02 m}$

Displacement, $y=0.0200\sin \left(\frac{\pi }{2}x\right)$

  $0.00\text{m}\le x\le 2.00\text{m}$

Formula Used:

Average value of kinetic energy per unit length:

  $\frac{dK}{dx}=\frac{1}{2}\mu {\left(\frac{\partial y}{\partial t}\right)}^2$

Here, $\mu$ is the mass per unit length.

Wave equation of standing wave in fundamental mode:

  $y=A\sin \left(kx\right)\cos \left(\omega t\right)$

Calculations:

  $\frac{\partial y}{\partial t}=A\sin kx\left(-\omega \sin \omega t\right)$

For maxima, equate the derivative with zero.

  $\begin{array}{l}\sin \left(kx\right)=0\\ kx=0\Rightarrow x=0\\ \text{or}\\ kx=\pi \\ x=\frac{\pi }{2\pi /\lambda }\\ x=\frac{\lambda }{2}=\frac{\left(2l\right)}{2}\\ x=\frac{2\left(2.00\right)}{2}=2.00\text{m}\end{array}$

Conclusion:

Thus, the value of x for which the average value of the kinetic energy per unit length is the greatest is $0.0\text{m}\&2.00\text{m}$ .

(d)

To determine

To Find: The value of x for which the elastic potential energy per unit length has the maximum value.

Explanation of Solution

Given:

Length of the wire, $l=2.00\text{m}$

Tension in the wire, $T=40.0\text{N}$

Mass of the wire, $m=0.100\text{kg}$

At the midpoint, amplitude is $A=2.00\text{cm = 0}\text{.02 m}$

Displacement, $y=0.0200\sin \left(\frac{\pi }{2}x\right)$

  $0.00\text{m}\le x\le 2.00\text{m}$

Formula Used:

Average value of elastic potential energy per unit length:

  $\frac{dU}{dx}=\frac{1}{2}\mu {\left(\frac{\partial y}{\partial x}\right)}^2$

Here, $\mu$ is the mass per unit length.

Wave equation of standing wave in fundamental mode:

  $y=A\sin \left(kx\right)\cos \left(\omega t\right)$

Calculations:

  $\frac{\partial y}{\partial x}=Ak\cos kx\cos \omega t$

For maxima, equate the derivative with zero.

  $\begin{array}{l}\cos \left(kx\right)=0\\ kx=\frac{\pi }{2}\\ x=\frac{\pi /2}{2\pi /\lambda }\\ x=\frac{\lambda }{4}=\frac{2l}{4}\\ x=\frac{2\left(2.00\right)}{4}=1.0\text{m}\end{array}$

Conclusion:

Thus, the value of x for which the average value of the elastic potential energy per unit length is the greatest is at $1.0\text{m}$ .