Question

A source of sound and an observer are moving each with a speed of $10\%$ of the speed of sound in air. If the frequency of sound heard by the observer when they approach each other is $400\text{ Hz}$ more than the frequency of sound heard by the observer when they move away from each other, then the frequency of the source of sound is:

• A. $440\text{ Hz}$
• B. $770\text{ Hz}$
• C. $550\text{ Hz}$
• D. $990\text{ Hz}$

Hint:

When $v_o = v_s = u$, the ratio of approaching to receding frequency is:
$\frac{f_{\text{app}}}{f_{\text{rec}}} = \left(\frac{v+u}{v-u}\right)^2$.
Here, with $u=0.1v$, the ratio is $\left(\frac{1.1}{0.9}\right)^2 = \frac{121}{81}$.
This structural identity helps to quickly verify the common denominator $99$ in calculations.

Answer 1

The Correct Option is D

Step 1: Set up the two Doppler expressions.
Let $v$ be the speed of sound and $f$ the true source frequency. Both source and observer move at $0.1v$, so put $x = 0.1$. Approaching, the observer hears $f_1 = f\frac{1+x}{1-x}$, and receding, he hears $f_2 = f\frac{1-x}{1+x}$.
Step 2: Combine into one fraction using a difference of squares.
Subtracting and putting over a common denominator, the numerator collapses to a difference of squares: \[ f_1 - f_2 = f\left[\frac{(1+x)^2-(1-x)^2}{(1-x)(1+x)}\right] = f\cdot\frac{4x}{1-x^2} \]
Step 3: Substitute the known values. \[ 400 = f\cdot\frac{4(0.1)}{1-(0.1)^2} = f\cdot\frac{0.4}{0.99} \]
Step 4: Solve for the source frequency. \[ f = \frac{400 \times 0.99}{0.4} \] \[ \boxed{f = 990\text{ Hz}} \]