Chapter 17: Q111P (page 513)

Question: A listener at rest (with respect to the air and the ground) hears a signal of frequency $f_{1}$ from a source moving toward him with a velocity of 15 m / s, due east. If the listener then moves toward the approaching source with a velocity of $25 \frac{m}{s}$ , due west, he hears a frequency $f_{2}$ that differs from $f_{1}$ by 37 Hz. What is the frequency of the source? (Take the speed of sound in air to be $340 \frac{m}{s}$ ).

Short Answer

Expert verified

The frequency of sound wave is 481 Hz.

Step by step solution

The given data

i)The speed of the source is $V_{S} = 15 \frac{m}{s}$
ii)The speed of the listener is $V_{D} = 25 \frac{m}{S}$
iii) The difference in frequency is $f_{2} - f_{1} = 37 H z$

Determine the concept of the Doppler Effect

Use the principle of the Doppler Effect and its corresponding equations to get the equations of $f_{1} and f_{2}$ in terms of f. Then we can use the value of the difference of $f_{1} and f_{2}$ and solve for the frequency of source f.

Formula:

The frequency received by the observer or the source according to Doppler’s Effect, (Consider detector at rest and source is approaching detector)

$f_{1} = f \frac{V}{V - V_{S}}$ (i)

(Consider source is approaching detector and detector is approaching source)

$f_{2} = f \times \frac{V + V_{D}}{V - V_{S}} (ii)$

Calculate frequency of sound wave

Consider the equation (i) of Doppler Effect for stationary detector and moving source as follows:

$f_{1} = f \frac{340}{340 - 15}$ ….. (1)

Now, the source as well asthedetector is moving towards each other. So, using equation (ii), we get

$f_{2} = f \times \frac{340 + 25}{340 - 15}$ ….. (2)

Now, we have the frequency difference as:

$f \times \frac{340 + 25}{340 - 15} - f \times \frac{340}{340 - 15} = 37 Hz f (\frac{340 + 25}{340 - 15} - \frac{340}{340 - 15}) = 37 Hz f (\frac{25}{325}) = 37 Hz f = 481 Hz$