Question

A source producing sound of frequency \(990\) Hz and an observer are initially at rest. If both the source and observer start moving simultaneously towards each other with accelerations \(2\ \mathrm{ms^{-2}}\) and \(4\ \mathrm{ms^{-2}}\) respectively, then the frequency of sound heard by the observer at a time \(t=5\) s is \[ (\text{Speed of sound in air}=340\ \mathrm{ms^{-1}}) \]

• A. \(960\) Hz
• B. \(1080\) Hz
• C. \(1050\) Hz
• D. \(900\) Hz

Hint:

For Doppler effect when the source and observer move towards each other, \[ \boxed{ f' = f \left( \frac{v+v_o}{v-v_s} \right). } \] If the bodies start from rest with uniform acceleration, \[ \boxed{v=at.} \]

Answer 1

The Correct Option is B

Step 1: Find the velocities of the source and observer. Initially both are at rest. Velocity of the observer after \[ 5\text{ s} \] is \[ v_o = at = 4\times5 = 20\text{ ms}^{-1}. \] Velocity of the source after \[ 5\text{ s} \] is \[ v_s = 2\times5 = 10\text{ ms}^{-1}. \]

Step 2: Apply the Doppler effect formula. For the source and observer moving towards each other, \[ f' = f \left( \frac{v+v_o}{v-v_s} \right). \] Substituting, \[ f' = 990 \left( \frac{340+20}{340-10} \right) = 990 \left( \frac{360}{330} \right). \] \[ = 990\times\frac{12}{11} = 1080\text{ Hz}. \]

Step 3: Write the answer. Hence, \[ \boxed{1080\text{ Hz}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.