Chapter 16, Problem 64P

(a)

To determine

The wave function for this vibration.

Explanation of Solution

Given:

Length of string is $2.00\text{m}$ .

Maximum amplitude is $3.00\text{cm}$ .

Frequency of the string is $100\text{Hz}$ .

Formula used:

Write expression for general form of wave function for 3^rd harmonic.

  $y_3\left(x,t\right)=A_3\sin \left(k_{3}x\right)\cos \left({\omega }_{3}t\right)$....... (1)

Write expression for angular velocity.

  ${\omega }_3=2\pi f_3$......... (2)

Write expression for wavelength of standing wave for string fixed at one end.

  ${\lambda }_3=\frac{4}{3}L$......... (3)

Write expression for wave number.

  $k_3=\frac{2\pi }{{\lambda }_3}$......... (4)

Calculation:

Substitute $100\text{Hz}$ for $f_3$ in equation (2).

  $\begin{array}{c}{\omega }_3=2\pi \left(100\text{Hz}\right)\\ =200\pi {\text{s}}^{\text{-1}}\end{array}$

Substitute $2\text{m}$ for $L$ in equation (3).

  $\begin{array}{c}{\lambda }_3=\frac{4}{3}\left(2\text{m}\right)\\ =\frac{8}{3}\text{m}\end{array}$

Substitute $\frac{8}{3}\text{m}$ for ${\lambda }_3$ in equation (4).

  $\begin{array}{c}{k}_3=\frac{2\pi }{8/3}\\ =\frac{3\pi }{4}\end{array}$

Substitute $3.00\text{cm}$ for $A_3$ , $200\pi {\text{s}}^{\text{-1}}$ for ${\omega }_3$ , $\frac{8}{3}\text{m}$ for ${\lambda }_3$ and $\frac{3\pi }{4}$ for $k_3$ in equation (1).

  $\begin{array}{l}{y}_3\left(x,t\right)=\left(3.00\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)\\ y_3\left(x,t\right)=\left(0.03\text{m}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)\end{array}$

Conclusion:

Thus, the wave function for vibration is $y_3\left(x,t\right)=\left(0.03\text{m}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)$ .

(b)

To determine

Function for kinetic energy; time at which kinetic energy is maximum.

Explanation of Solution

Given:

Length of string is $2.00\text{m}$ .

Maximum amplitude is $3.00\text{cm}$ .

Frequency of the string is $100\text{Hz}$ .

Formula used:

Write expression for general form of wave function for 3^rd harmonic.

  $y_3\left(x,t\right)=A_3\sin \left(k_{3}x\right)\cos \left({\omega }_{3}t\right)$......... (1)

Write expression for angular velocity.

  ${\omega }_3=2\pi f_3$......... (2)

Write expression for wavelength of standing wave for string fixed at one end.

  ${\lambda }_3=\frac{4}{3}L$......... (3)

Write expression for wave number.

  $k_3=\frac{2\pi }{{\lambda }_3}$......... (4)

Write expression for kinetic energy.

  $dK=\frac{1}{2}\mu v_y^{2}dx$......... (5)

Calculation:

Substitute $100\text{Hz}$ for $f_3$ in equation (2).

  $\begin{array}{c}{\omega }_3=2\pi \left(100\text{Hz}\right)\\ =200\pi {\text{s}}^{\text{-1}}\end{array}$

Substitute $2\text{m}$ for $L$ in equation (3).

  $\begin{array}{c}{\lambda }_3=\frac{4}{3}\left(2\text{m}\right)\\ =\frac{8}{3}\text{m}\end{array}$

Substitute $\frac{8}{3}\text{m}$ for ${\lambda }_3$ in equation (4).

  $\begin{array}{c}{k}_3=\frac{2\pi }{8/3}\\ =\frac{3\pi }{4}\end{array}$

Substitute $3.00\text{cm}$ for $A_3$ , $200\pi {\text{s}}^{\text{-1}}$ for ${\omega }_3$ , $\frac{8}{3}\text{m}$ for ${\lambda }_3$ and $\frac{3\pi }{4}$ for $k_3$ in equation (1).

  $\begin{array}{l}{y}_3\left(x,t\right)=\left(3.00\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)\\ y_3\left(x,t\right)=\left(0.03\text{m}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)\end{array}$

Differentiate above expression with respect to $t$ .

  $\begin{array}{c}\frac{d\left(y_3\left(x,t\right)\right)}{dt}=\frac{d}{dt}\left[\left(0.03\text{m}\right)\sin \left(\frac{3\pi }{4}x\right)\cos \left(200\pi {\text{s}}^{\text{-1}}t\right)\right]\\ \frac{d\left(y_3\left(x,t\right)\right)}{dt}=-\left(200\pi {\text{s}}^{\text{-1}}\right)\left(0.03\text{m}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\\ \frac{d\left(y_3\left(x,t\right)\right)}{dt}=-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\end{array}$

Substitute $v_y$ for $\frac{d\left(y_3\left(x,t\right)\right)}{dt}$ in above expression.

  $v_y=-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)$

Substitute $-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)$ for $v_y$ in equation (5).

  $\begin{array}{l}dK=\frac{1}{2}\mu {\left(-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\right)}^{2}dx\\ dK=\frac{1}{2}{\left(-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\right)}^2\mu dx\end{array}$

Write expression for condition of maximum kinetic energy.

  $\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)=1$

Solve above expression.

  $\begin{array}{c}\left(200\pi {\text{s}}^{\text{-1}}\right)t=\frac{\pi }{2},\frac{3\pi }{2},\mathrm{.....}\\ t=\frac{1}{200\pi {\text{s}}^{\text{-1}}}\frac{\pi }{2},\frac{1}{200\pi {\text{s}}^{\text{-1}}}\frac{3\pi }{2},\mathrm{...}\\ t=0.0025\text{s,0}\text{.0075}\text{s,}\mathrm{...}\end{array}$

Conclusion:

Thus the function for kinetic energy is $dK=\frac{1}{2}{\left(-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\right)}^2\mu dx$ ; time at which kinetic energy is maximum are $0.0025\text{s,0}\text{.0075}\text{s,}\mathrm{...}$ .

(c)

To determine

The maximum kinetic energy of string by integration.

Explanation of Solution

Given:

The function for kinetic energy is $dK=\frac{1}{2}{\left(-\left(6\pi \text{m}/\text{s}\right)\sin \left(\frac{3\pi }{4}x\right)\sin \left(200\pi {\text{s}}^{\text{-1}}t\right)\right)}^2\mu dx$

Formula used:

Write expression for general form of wave function for 3^rd harmonic.

  $y_3\left(x,t\right)=A_3\sin \left(k_{3}x\right)\cos \left({\omega }_{3}t\right)$......... (1)

Write expression for kinetic energy.

  $dK=\frac{1}{2}\mu v_y^{2}dx$......... (2)

Calculation:

Differentiate equation (1) with respect to $t$ .

  $\begin{array}{c}\frac{d}{dx}\left(y_3\left(x,t\right)\right)=\frac{d}{dx}\left[A_3\sin \left(k_{3}x\right)\cos \left({\omega }_{3}t\right)\right]\\ v_y=-{\omega }_3{A}_3\sin \left(k_{3}x\right)\sin \left({\omega }_{3}t\right)\end{array}$

Substitute $-{\omega }_3{A}_3\sin \left(k_{3}x\right)\sin \left({\omega }_{3}t\right)$ for $v_y$ in equation (2).

  $dK=\frac{1}{2}\mu {\left({\omega }_3{A}_3\sin \left(k_{3}x\right)\sin \left({\omega }_{3}t\right)\right)}^{2}dx$

Integrate above expression from $0$ to $L$ .

  $\begin{array}{c}{\displaystyle \int dK}={\displaystyle \underset{0}{\overset{L}{\int }}\left[\frac{1}{2}\mu {\left({\omega }_3{A}_3\sin \left(k_{3}x\right)\sin \left({\omega }_{3}t\right)\right)}^2\right]dx}\\ K_{\max }=\frac{1}{4}m{\omega }^2{A}^2\end{array}$

Conclusion:

Thus, the value of maximum kinetic energy is $K_{\max }=\frac{1}{4}m{\omega }^2{A}^2$ .