A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him?

The correct option is 750Hz

When the source is moving towards the observer, the apparent frequency is given by:

fapp = (v/(v - vs)) * fsource

where:

• fapp is the apparent frequency
• v is the speed of sound (assumed to be 350 m/s)
• vs is the velocity of the source (50 m/s)
• fsource is the actual frequency of the source (1000 Hz)

Substituting the values, we get:

1000 = (350 / (350 - 50)) * fsource

fsource = 1000 * (300 / 350) = 857.14 Hz (approximately)

When the source is moving away from the observer, the apparent frequency is given by:

fapp = (v / (v + vs)) * fsource

Substituting the values, and using the actual frequency we calculated above, we get:

fapp = (350 / (350 + 50)) * 857.14

However, the question provides the apparent frequency when approaching as 1000Hz. Let's use this value to find the actual frequency of the source and then calculate the frequency when moving away:

1000 = (350 / (350 - 50)) * fsource

fsource = (1000 * (350 - 50)) / 350 = (1000 * 300) / 350 = 857.14 Hz (approximately)

Now let's calculate the apparent frequency when the source is moving away:

fapp = (350 / (350 + 50)) * 857.14 ≈ 750 Hz

Therefore, the apparent frequency of the source when it is moving away from the observer is approximately 750 Hz.