Home
Class 12
PHYSICS
A sound source is moving with speed 50 m...

A sound source is moving with speed `50 m//s` towards a fixed observer. Frequency observed by observer is `1000Hz`. Find out apparent frequency observed by observer when source is moving away from observer (Speed of sound `=350 m//s`)

A

750 Hz

B

857 Hz

C

1143 Hz

D

807 Hz

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will use the Doppler effect formula for sound. The frequency observed by a stationary observer when the source is moving towards or away from them can be calculated using the following formulas: 1. When the source is moving towards the observer: \[ f' = f \frac{v}{v - v_s} \] where: - \( f' \) = observed frequency - \( f \) = actual frequency of the source - \( v \) = speed of sound - \( v_s \) = speed of the source 2. When the source is moving away from the observer: \[ f'' = f \frac{v}{v + v_s} \] where \( f'' \) is the observed frequency when the source is moving away. ### Step-by-Step Solution **Step 1: Identify the known values.** - Speed of sound, \( v = 350 \, \text{m/s} \) - Speed of the source, \( v_s = 50 \, \text{m/s} \) - Observed frequency when the source is approaching, \( f' = 1000 \, \text{Hz} \) **Step 2: Use the first formula to find the actual frequency \( f \).** Using the formula for the frequency when the source is moving towards the observer: \[ f' = f \frac{v}{v - v_s} \] Substituting the known values: \[ 1000 = f \frac{350}{350 - 50} \] \[ 1000 = f \frac{350}{300} \] \[ 1000 = f \cdot \frac{7}{6} \] Now, solve for \( f \): \[ f = 1000 \cdot \frac{6}{7} = \frac{6000}{7} \approx 857.14 \, \text{Hz} \] **Step 3: Use the second formula to find the apparent frequency \( f'' \) when the source is moving away.** Using the formula for the frequency when the source is moving away from the observer: \[ f'' = f \frac{v}{v + v_s} \] Substituting the values we have: \[ f'' = \left(\frac{6000}{7}\right) \frac{350}{350 + 50} \] \[ f'' = \left(\frac{6000}{7}\right) \frac{350}{400} \] \[ f'' = \left(\frac{6000}{7}\right) \cdot \frac{7}{8} = \frac{6000}{8} = 750 \, \text{Hz} \] ### Final Answer: The apparent frequency observed by the observer when the source is moving away from them is \( 750 \, \text{Hz} \).
Promotional Banner

Topper's Solved these Questions

  • JEE MAINS

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise Chemistry|1 Videos
  • JEE MAIN

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise All Questions|452 Videos
  • JEE MAINS 2020

    JEE MAINS PREVIOUS YEAR ENGLISH|Exercise PHYSICS|250 Videos