Chapter 12, Problem 58P

Solution

(a)

To Calculate: The frequency that an observer can hear, who is located at rest in due north.

The velocity component does not get affected by blowing wind, so the frequency that can be heard is $570\text{ Hz}$ in due north.

  $$

Given:

Frequency of whistle $\left(f_o\right)$

  $=570\text{ Hz}.$

Wind velocity from north $\left(v_w\right)$

  $=12\text{ m/s}.$

Formula used:

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $v_{whistle}=$ speed of whistle sound

  $f=$ frequency in due north.

  $f_0=$ frequency of whistle.

Calculation:

The velocity component of the sound  v , reaching the observer will not be affected by the blowing wind.

The frequency that can be heard by the observer is,

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $f=f_o$

Therefore, the frequency that can be heard is $570\text{ Hz}.$

Conclusion:

The velocity component does not affect by blowing wind, so the frequency that can be heard is $570\text{ Hz}$ in due north.

(b)

To Calculate: The frequency that an observer can hear, who is located at rest in due south.

The velocity component does not get affected by blowing wind, so the frequency that can be heard is $570\text{ Hz}$ in due south.

Given:

Frequency of whistle $\left(f_o\right)$

  $=570\text{ Hz}.$

Wind velocity from north $\left(v_w\right)$

  $=12\text{ m/s}.$

Formula used:

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $v_{whistle}=$ speed of whistle sound

  $v_{sound}=$ Speed of sound

  $f=$ frequency in due north

  $f_0=$ frequency of whistle

Calculation:

There is no Doppler shift and the frequency that can be heard by the observer is,

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $f=f_o$

Therefore, the frequency that can be heard is $570\text{ Hz}.$

Conclusion:

The velocity component does not affect by blowing wind, so the frequency that can be heard is $570\text{ Hz}$ in due south.

(c)

To Calculate: The frequency that an observer can hear who is located at rest in due east.

  $570\text{ Hz}.$

Given:

Frequency of whistle, $\left(f_o\right)$

  $=570\text{ Hz}.$

Wind velocity from north, $\left(v_w\right)$

  $=12\text{ m/s}.$

Formula used:

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $v_{whistle}=$ speed of whistle sound

  $v_{sound}=$ Speed of sound

  $f=$ frequency in due north

  $f_0=$ frequency of whistle

Calculation:

There is no Doppler shift and the frequency that can be heard by the observer is,

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $f=f_o$

Therefore, the frequency that can be heard is $570\text{ Hz}.$

Conclusion:

The velocity component does not affect by blowing wind, so the frequency that can be heard is $570\text{ Hz}$ in due east.

(d)

To Calculate: The frequency that an observer can hear who is located at rest is due west of the whistle.

570 Hz.

Given:

Frequency of whistle, $\left(f_o\right)$

  $=570\text{ Hz}.$

Wind velocity from north, $\left(v_w\right)$

  $=12\text{ m/s}.$

Formula used:

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $v_{whistle}=$ speed of whistle sound

  $v_{sound}=$ Speed of sound

  $f=$ frequency in due north

  $f_0=$ frequency of whistle

Calculation:

The frequency that can be heard by the observer is,

  $f=f_o\left(\frac{v_{whistle}}{v_{sound}}\right)$

  $f=f_o$

Therefore, the frequency that can be heard is $570\text{ Hz}.$

Conclusion:

The frequency that can be heard is $570\text{ Hz}.$ due west.

(e)

To Calculate: The frequency that can be heard by cyclist heading north.

  $587.1\text{ Hz}$

Given:

Frequency of whistle, $\left(f_o\right)$

  $=570\text{ Hz}.$

Wind velocity from north, $\left(v_w\right)$

  $=12\text{ m/s}.$

Velocity of the cyclist towards North, $v_{cyclist}=15.0\text{m/s}$

Formula used:

  $f=f_o\left(\frac{v_s+v_w+v_{cyclist}}{v_s+v_w}\right)$

  $v_s=$ speed of whistle sound

  $v_w=$ Speed of wind

  $v_{cyclist}=$ speed of cyclist

  $f=$ frequency

  $f_0=$ frequency of sound (whistle)

Calculation:

The frequency heard by the observer is,

  $f=f_o\left(\frac{v_s+v_w+v_{cyclist}}{v_s+v_w}\right)$

Where,

  $v_s$ = velocity of sound, 343 m/s

Substituting the values,

  $\begin{array}{c}f=570\left(\frac{343+15+12}{343+12}\right)\text{ Hz}\\ \text{=1}\text{.033}\left(570\right)\text{Hz}\\ =588.81\text{Hz}\end{array}$

Conclusion:

The frequency that can be heard is $588.1\text{ Hz}$ when a cyclist heading to the north.

(f)

To Calculate: The frequency that can be heard, when heading towards the whistle.

  $f=594.92\text{ Hz}$

Given:

Frequency of whistle, $\left(f_o\right)$

  $=570\text{ Hz}.$

Velocity of the cyclist towards North, $v_{cyclist}=15.0\text{m/s}$

Formula used:

  $f=f_o\left(\frac{v_s+v_{cyclist}}{v_s}\right)$

  $v_s=$ speed of whistle sound

  $v_{cyclist}=$ speed of cyclist

  $f=$ frequency.

  $f_0=$ frequency of whistle, Hz.

Calculation:

The velocity component of the sound  v , which is reaching the observer will be unaffected by the blowing wind.

There is no Doppler shift and the frequency that can be heard by the observer is,

  $f=f_o\left(\frac{v_s+v_{cyclist}}{v_s}\right)$

Substituting the values,

  $f^1=570\left(\frac{15+343}{343}\right)\text{ Hz}$

  $f^1=594.92\text{ Hz}$

Conclusion:

The frequency that can be heard, when heading towards the whistle at a speed of $15\text{ m/s}$ due west is $594.92\text{ Hz}$ .