Chapter 13, Problem 45QAP

To determine

(a)

The value of amplitude of pressure wave.

Answer to Problem 45QAP

  $A=1.0\text{atm}$

Explanation of Solution

Given:

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Formula used:

  $y(x,t)=A\cos (kx-\omega t)$

Calculation:

The displacement of transverse wave is given by,

  $p(x,t)=A\cos (kx-\omega t)$

Comparing above equation with the give equation,

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Amplitude, A= 0.5 atm

Conclusion:

Amplitude of the pressure wave is 0.5atm.

To determine

(b)

The value of wave number of given pressure wave.

Answer to Problem 45QAP

  $k=6.0\text{rad/m}$

Explanation of Solution

Given:

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Formula used:

  $y(x,t)=A\cos (kx-\omega t)$

Calculation:

The displacement of transverse wave is given by,

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Comparing above equation with the give equation,

  $y(x,t)=A\cos (kx-\omega t)$

  $k=6.0\text{rad/m}$

Conclusion:

Wave number of the pressure wave is $k=6.0\text{rad/m}$

To determine

(c)

The value of frequency of pressure wave.

Answer to Problem 45QAP

  $f=0.6369\text{Hz}$

Explanation of Solution

Given:

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Formula used:

  $y(x,t)=A\cos (kx-\omega t)$

Calculation:

The displacement of transverse wave is given by,

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Comparing above equation with the give equation,

  $y(x,t)=A\cos (kx-\omega t)$

We get,

  $\begin{array}{l}\omega =4\text{rad/s}\\ 2\pi f=4\\ f=\frac{2}{3.14}\\ f=0.6369\text{Hz}\end{array}$

Conclusion:

Frequency of the pressure wave is $f=0.6369\text{Hz}$.

To determine

(d)

The value of speed of pressure wave.

Answer to Problem 45QAP

  $v_p=0.7287\text{m/s}$

Explanation of Solution

Given:

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Formula used:

  $y(x,t)=A\cos (kx-\omega t)$

Calculation:

The displacement of transverse wave is given by,

  $p(x,t)=(1.0\text{atm})\cos (6.0x-4.0t)$

Comparing above equation with the give equation,

  $y(x,t)=A\cos (kx-\omega t)$

We get,

  $\begin{array}{l}k=\frac{2\pi }{\lambda }=6\\ \text{but,}{v}_p=\lambda f\\ \therefore \lambda =\frac{v_p}{f}\\ \therefore 6=\frac{2\pi }{\frac{v_p}{f}}=\frac{2\pi f}{v_p}\\ 6=\frac{2\pi f}{v_p}\\ v_p=\frac{2\pi (0.6963)}{6}\\ v_p=0.7287\text{m/s}\end{array}$

Conclusion:

Speed of the pressure wave is $v_p=0.7287\text{m/s}$.