A stationary source is emitting sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f0. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms

Engineering

Physics

Approximation and Order of Magnitude

Straight Line Motion

Dopplers Effect

Question

A stationary source is emitting sound at a fixed frequency f₀, which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f₀. What is the difference in the speeds of the cars (in km per hour) to the nearest integer? The cars are moving at constant speeds much smaller than the speed of sound which is 330 ms^–1.

$ρ_{1} = \frac{c + v_{1}}{c - v_{1}} ρ_{0}$

$ρ_{2} = \frac{c + v_{2}}{c - v_{2}} ρ_{0}$

$\frac{Δρ}{ρ_{0}} = (\frac{c + v_{1}}{c - v_{1}}) - (\frac{c + v_{2}}{c - v_{2}}) = \frac{1 .2}{100}$

$= \frac{c^{2} - v_{1} v_{2} + {cv}_{1} - {cv}_{2} - (c^{2} - v_{1} v_{2} - {cv}_{1} + {cv}_{2})}{(c - v_{1}) (c - v_{2})}$

$= \frac{2 c (v_{1} - v_{2})}{c^{2}} = \frac{1 .2}{100}$

$= v_{1} - v_{2} = \frac{1 .2}{100} \times \frac{330}{2} \times \frac{18}{5}$

$= \frac{1 .2 \times 33 \times 18}{100} \approx 7 km / hr .$

This problem involves the Doppler effect for sound waves, where a stationary source emits sound that is reflected by moving objects (cars). The frequency heard by the moving car and then the frequency reflected back to the source (or observed) changes due to the motion of the car.

Let's break it down step by step:

Step 1: Frequency observed by a moving car
The source is stationary, emitting frequency $f_{0}$ . A car is moving towards the source with speed $v$ . The frequency observed by the car, $f_{1}$ , is given by the Doppler effect formula for a moving observer approaching a stationary source: $f_{1} = f_{0} (\frac{V + v}{V})$ where $V$ is the speed of sound (330 m/s).

Step 2: Frequency of sound reflected from the car
The car now acts as a moving source, reflecting sound of frequency $f_{1}$ back towards the stationary original source. For a moving source approaching a stationary observer (the original source), the frequency observed, $f_{2}$ , is: $f_{2} = f_{1} (\frac{V}{V - v})$

Step 3: Combined effect for the reflected frequency
Substitute the expression for $f_{1}$ into the equation for $f_{2}$ : $f_{2} = f_{0} (\frac{V + v}{V}) (\frac{V}{V - v}) = f_{0} (\frac{V + v}{V - v})$

This is the frequency of the sound reflected from a car moving with speed $v$ .

Step 4: Apply the approximation
The problem states that the car speeds are much smaller than the speed of sound ( $v ≪ V$ ). We can use the binomial approximation to simplify the expression. For small $x$ , $\frac{1 + x}{1 - x} \approx 1 + 2 x$ when $x ≪ 1$ . Here, $x = \frac{v}{V}$ . Therefore: $f_{2} \approx f_{0} [1 + 2 (\frac{v}{V})] = f_{0} + 2 f_{0} \frac{v}{V}$

Step 5: Relate to the problem
For two cars with speeds $v_{1}$ and $v_{2}$ , the difference in their reflected frequencies is: $Δ f_{2} = f_{2} (v_{1}) - f_{2} (v_{2}) \approx 2 f_{0} \frac{v_{1} - v_{2}}{V} = 2 f_{0} \frac{Δ v}{V}$ where $Δ v = | v_{1} - v_{2} |$ is the difference in their speeds.

We are told this frequency difference is 1.2% of $f_{0}$ : $Δ f_{2} = 0.012 f_{0}$

Therefore: $2 f_{0} \frac{Δ v}{V} = 0.012 f_{0}$

We can cancel $f_{0}$ from both sides: $2 \frac{Δ v}{V} = 0.012$

Step 6: Solve for the speed difference
Rearranging the equation: $Δ v = \frac{0.012 \times V}{2} = \frac{0.012 \times 330}{2}$

First, calculate the numerator: $0.012 \times 330 = 3.96$
Then, divide by 2: $Δ v = \frac{3.96}{2} = 1.98 m / s$

Step 7: Convert to km/h
To convert from m/s to km/h, multiply by $\frac{18}{5} = 3.6$ : $Δ v = 1.98 \times 3.6 = 7.128 k m / h$

Rounding to the nearest integer gives 7 km/h.

Related Concepts and Formulae

Doppler Effect for Sound: The change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. The general formula for the observed frequency when the source and observer are moving along the line joining them is: $f_{o b s} = f_{s o u r c e} (\frac{V \pm v_{o b s}}{V \mp v_{s o u r c e}})$ The signs are chosen such that the frequency increases when the source and observer move towards each other.

Binomial Approximation: A useful mathematical tool for simplifying expressions where a term is much smaller than 1. For $x ≪ 1$ , $(1 \pm x)^{n} \approx 1 \pm n x$ .

Detailed Solution

Related Concepts and Formulae

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