Question

Two observers \(A\) and \(B\) are moving towards a stationary source of sound with speeds \(0.8v_0\) and \(0.6v_0\) respectively, where \(v_0\) is the speed of sound in air. The ratio of the frequencies of the sound heard by the observers \(A\) and \(B\) is

• A. \(4:3\)
• B. \(9:8\)
• C. \(1:1\)
• D. \(2:3\)

Hint:

For a moving observer and a stationary source, \[ \boxed{ f' = f\left(\frac{v\pm v_o}{v}\right). } \] Use the positive sign when the observer moves towards the source and the negative sign when moving away.

Answer 1

The Correct Option is B

Step 1: Use the Doppler effect formula for a moving observer. For a stationary source and an observer moving towards the source, \[ f' = f\left(\frac{v+v_o}{v}\right), \] where \[ v=v_0 \] is the speed of sound.

Step 2: Find the observed frequencies. For observer \(A\), \[ v_A=0.8v_0, \] so \[ f_A = f\left(\frac{v_0+0.8v_0}{v_0}\right) = 1.8f. \] For observer \(B\), \[ v_B=0.6v_0, \] thus \[ f_B = f\left(\frac{v_0+0.6v_0}{v_0}\right) = 1.6f. \]

Step 3: Calculate the ratio. Therefore, \[ \frac{f_A}{f_B} = \frac{1.8}{1.6} = \frac{18}{16} = \frac98. \] Hence, \[ \boxed{f_A:f_B=9:8.} \] Therefore, the correct option is \(\boxed{(B)}\).