Q45E - University Physics with Modern Physics

Question

On the planet Arrakis a male ornithoid is flying toward his mate at $\mathbf{25}\mathbf{m}/\mathbf{s}$ while singing at a frequency of  $\mathbf{1200}\mathbf{Hz}$ . If the stationary female hears a tone of  $\mathbf{1240}\mathbf{Hz}$, what is the speed of sound in the atmosphere of Arrakis?

Step-by-Step Solution

Answer

Sound travels at a speed of  $775\text{\hspace{0.17em}m}/\text{s}$ .

Step 1: Given data

Speed of the male ornithoid is   $\mathbf{2}5\mathbf{m}/\mathbf{s}$

Frequency while singing by male is $1200\text{\hspace{0.17em}\hspace{0.17em}Hz}$  

Frequency heard by the female is $1240\text{\hspace{0.17em}\hspace{0.17em}Hz}$ 

Step 2: Doppler’s effect

The Doppler’s effect is given by the formula, 

${\mathbf{f}}_L\mathbf{=}\frac{\mathbf{v}\mathbf{+}{\mathbf{v}}_L}{\mathbf{v}\mathbf{+}{\mathbf{v}}_S}{\mathbf{f}}_S$ 

where

• ${\mathit{f}}_{\mathbf{L}}$  is the frequency of the wave that is heard by the listener.
• $\mathit{v}$  is the wave speed.
• ${\mathit{v}}_{\mathbf{L}}$  is the listener’s speed.
• ${\mathit{v}}_{\mathbf{s}}$  is the sound source’s speed.
• ${\mathbf{f}}_{\mathbf{s}}$  is the sound source’s frequency.

Step 3: Consider the male is moving towards the listener

$\begin{array}{c}{f}_L=\frac{v+v_L}{v+v_S}{f}_S\\ 1240\text{\hspace{0.17em}Hz}=\frac{v+0}{v-25\text{\hspace{0.17em}m}/\text{s}}\times 1200\text{\hspace{0.17em}Hz}\\ \frac{v}{v-25\text{\hspace{0.17em}m}/\text{s}}=\frac{31}{30}\\ v=775\text{\hspace{0.17em}m/s}\end{array}$

So, the speed of the sound in the atmosphere is $775\text{\hspace{0.17em}m/s}$.