Question

For sound waves in air with frequency 1000 Hz a displacement amplitude of produces a pressure amplitude of $3.0 \times 10^{ - 2}\; {\rm{ Pa}}$ . Use $v_{\rm sound}$ = 344 ${\rm m/s}$ .

For sound waves in air with frequency 1000 Hz a displacement amplitude of x10^ - 8 m produces a pressure amplitude of 3.0 x 10^ - 2Pa. Use v_sound= 344 m/s.a) What is the wavelength of these waves?b) For 1000 - Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? Express your answer using two significant figures. c) For what wavelength will waves with a displacement amplitude of 1.2 x 10^ - 8m produce a pressure amplitude of 1.5 x 10^ - 3Pa? Express your answer using two significant figures. d) For what frequency will waves with a displacement amplitude of 1.2 x10^ - 8 m produce a pressure amplitude of 1.5 x 10^ - 3Pa? Express your answer using two significant figures.

a) What is the wavelength of these waves?

b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa?

Express your answer using two significant figures.

c) For what wavelength will waves with a displacement amplitude of $1.2 \times 10^{ - 8}\; {\rm{ m}}$ produce a pressure amplitude of $1.5 \times 10^{ - 3}\; {\rm{ Pa}}?$

Express your answer using two significant figures.

d) For what frequency will waves with a displacement amplitude of $1.2 \times 10^{ - 8}\; {\rm{ m}}$ produce a pressure amplitude of $1.5 \times 10^{ - 3}\; {\rm{ Pa}}?$

Express your answer using two significant figures.

Answer 1

Concepts and reason

The concepts required to solve this problem is wave number, wave speed, and pressure amplitude equation of the sound wave.

Initially, calculate the wavelength of the wave by using the expression of the wave speed. Calculate the wavenumber of the wave and then calculate the value of the bulk modulus by using the pressure amplitude equation of the wave. After that, calculate the displacement amplitude of the wave by using the pressure amplitude equation of the wave.

Later, calculate the wavelength of the wave by using the expression for wave number.

Finally, calculate the frequency of the wave by using wave speed.

Fundamentals

The expression for wave speed is,

$v = \lambda f$

Here, f is the frequency and $\lambda$ is the wavelength.

The expression for wave number is,

$k = \frac{\omega }{v}$

Here, $\omega$ is the angular frequency and v is the speed of the sound wave.

The expression for angular frequency is,

$\omega = 2\pi f$

Here, f is the frequency.

The expression for the pressure amplitude of the wave is,

$P = BkA$

Here, B is the bulk modulus, k is the wave number, and A is the displacement amplitude of the wave.

(a)

The expression for wave speed is,

$v = \lambda f$

Rewrite the above expression for $\lambda$.

$\lambda = \frac{v}{f}$

Substitute 344 m/s for v and 1,000 Hz for f in the above equation.

$\begin{array}{c}\\\lambda = \frac{{344\;{\rm{m/s}}}}{{1,000\;{\rm{Hz}}}}\\\\ = 0.344\;{\rm{Hz}}\\\end{array}$

Therefore, the wavelength of these waves is 0.344 Hz.

(b)

The expression for wave number is,

$k = \frac{\omega }{v}$

The expression for angular frequency is,

$\omega = 2\pi f$

Substitute $2\pi f$ for $\omega$ in the equation $k = \frac{\omega }{v}$.

$k = \frac{{2\pi f}}{v}$

Substitute 344 m/s for v and 1,000 Hz for f in the above equation.

$\begin{array}{r}\\k = \frac{{2\left( {3.14} \right)\left( {1,000\;{\rm{Hz}}} \right)}}{{\left( {344\;{\rm{m/s}}} \right)}}\\\\ = 18.25\;{\rm{rad/m}}\\\end{array}$

The expression for the pressure amplitude of the wave is,

$P = BkA$

Rewrite the above expression for B.

$B = \frac{P}{{kA}}$

Substitute $3.0 \times {10^{ - 2}}\;{\rm{Pa}}$ for P , 18.25 rad/m for k, and $1.2 \times {10^{ - 8}}\;{\rm{m}}$ for A in the above equation.

$\begin{array}{c}\\B = \frac{{3.0 \times {{10}^{ - 2}}\;{\rm{Pa}}}}{{\left( {18.25\;{\rm{rad/m}}} \right)\left( {\;1.2 \times {{10}^{ - 8}}\;{\rm{m}}} \right)}}\\\\ = 1.369 \times {10^5}\;{\rm{Pa}}\;\\\end{array}$

Calculate the displacement amplitude of the wave as follows:

$P = BkA$

Rewrite the above expression for A.

$A = \frac{P}{{Bk}}$

Substitute $3.0 \times {10^{ - 2}}\;{\rm{Pa}}$30 Pa for P , 18.25 rad/m for k, and $1.369 \times {10^5}\;{\rm{Pa}}$ for B in the above equation.

$\begin{array}{c}\\A = \frac{{3.0 \times {{10}^{ - 2}}\;{\rm{Pa}}}}{{\left( {1.369 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {18.25\;{\rm{rad/m}}} \right)}}\\\\ = 1.2 \times {10^{ - 5}}\;{\rm{m}}\\\end{array}$

Therefore, the displacement amplitude of the wave is $1.2 \times {10^{ - 5}}\;{\rm{m}}$.

(c)

Calculate the wave number as follows:

$P = BkA$

Rewrite the above expression for k.

$k = \frac{P}{{BA}}$

Substitute $1.5 \times {10^{ - 3}}\;{\rm{Pa}}$ for P, $1.369 \times {10^5}\;{\rm{Pa}}$ for B, and $1.2 \times {10^{ - 8}}\;{\rm{m}}$ for A in the above equation.

$\begin{array}{c}\\k = \frac{{\left( {1.5 \times {{10}^{ - 3}}\;{\rm{Pa}}} \right)}}{{\left( {1.369 \times {{10}^5}\;{\rm{Pa}}} \right)\left( {1.2 \times {{10}^{ - 8}}\;{\rm{m}}} \right)}}\\\\ = 0.91\;{\rm{rad/m}}\\\end{array}$

The expression for wavelength is,

$\lambda = \frac{{2\pi }}{k}$

Substitute 0.91 rad/m in the above equation.

$\begin{array}{c}\\\lambda = \frac{{2\left( {3.14} \right)}}{{\left( {0.91\;{\rm{rad/m}}} \right)}}\\\\ = 6.9\;{\rm{m}}\\\end{array}$

Therefore, the wavelength is 6.9 m.

(d)

The expression for frequency is,

$f = \frac{v}{\lambda }$

Substitute 344 m/s for v and 6.9 m for $\lambda$ in the above equation.

$\begin{array}{c}\\f = \frac{{344\;{\rm{m/s}}}}{{6.9\;{\rm{m}}}}\\\\ = 49.85\\\\ \approx 50\;{\rm{Hz}}\\\end{array}$

Therefore, the required frequency is 50 Hz.

Ans: Part a

Therefore, the wavelength of these waves is 0.344 Hz.

Part b

Therefore, the displacement amplitude of the wave is $1.2 \times {10^{ - 5}}\;{\rm{m}}$.

Part c

Therefore, the wavelength is 6.9 m.

Part d

Therefore, the required frequency is 50 Hz.