Question

A sound source of 1000 Hz frequency approaches the observer with speed \( 20~\text{ms}^{-1} \). The observed frequency of sound is nearly: (speed of sound in air \( = 340~\text{ms}^{-1} \))

• A. 1060 Hz
• B. 940 Hz
• C. 1020 Hz
• D. 1000 Hz

Hint:

When a sound source moves towards you, the denominator must decrease (\( v - v_s \)), which naturally causes the fraction to be greater than 1, resulting in a higher apparent pitch. Keeping track of this physical behavior helps prevent mixing up sign conventions.

Answer 1

The Correct Option is A

Step 1: Recall the Doppler effect for sound.
When a sound source moves toward a still listener, the listener hears a higher pitch. The apparent frequency is: \[ f' = f\left(\frac{v}{v - v_s}\right) \] where $v$ is the speed of sound and $v_s$ is the source speed.
Step 2: List the given values.
Source frequency $f = 1000$ Hz, source speed $v_s = 20\ ms^{-1}$, speed of sound $v = 340\ ms^{-1}$.
Step 3: Put the values in.
Because the source moves toward the listener we subtract $v_s$ in the bottom: \[ f' = 1000\left(\frac{340}{340 - 20}\right) = 1000\left(\frac{340}{320}\right) \]
Step 4: Simplify the fraction.
\[ \frac{340}{320} = \frac{17}{16} = 1.0625 \]
Step 5: Multiply out.
\[ f' = 1000 \times 1.0625 = 1062.5\ \text{Hz} \]
Step 6: Round to the nearest option.
\[ \boxed{f' \approx 1060\ \text{Hz}} \]