Question

An observer, moving in a straight line with velocity \(100\ \text{m s}^{-1}\), passes a stationary observer. If the object emits note of \(400\) Hz while moving, the change in the frequency noticed by the observer as the object passes past him is \((\text{speed of sound in air}=300\ \text{m s}^{-1})\)

• A. \(350\) Hz
• B. \(300\) Hz
• C. \(200\) Hz
• D. \(100\) Hz
• E. \(150\) Hz

Hint:

In Doppler effect, the jump in frequency when the source passes the observer can be found by subtracting the receding value from the approaching value.

Answer 1

The Correct Option is B

For a moving source and stationary observer: When approaching: \[ f_1=f\frac{v}{v-v_s} =400\cdot \frac{300}{300-100} =400\cdot \frac{300}{200}=600\text{ Hz} \] When receding: \[ f_2=f\frac{v}{v+v_s} =400\cdot \frac{300}{300+100} =400\cdot \frac{300}{400}=300\text{ Hz} \] Change in observed frequency: \[ \Delta f=f_1-f_2=600-300=300\text{ Hz} \] Hence, \[ \boxed{(B)\ 300\text{ Hz}} \]