Question

Two cars X and Y are approaching each other with velocities 36 km/h and 72 km/h respectively. The frequency of a whistle sound as emitted by a passenger in car X, heard by the passenger in car Y is 1320 Hz. If the velocity of sound in air is 340 m/s, the actual frequency of the whistle sound produced is _________ Hz.

Hint:

To remember the signs in the Doppler formula, think about the physical effect. When source and observer move towards each other, the frequency should increase, so the fraction must be greater than 1. When they move away, the frequency should decrease, so the fraction must be less than 1. Adjust the signs of \(v_{observer}\) and \(v_{source}\) accordingly.

Answer 1

Correct Answer: 1210

Step 1: Understanding the Question:
This is a classic Doppler effect problem. The source (car X) and the observer (car Y) are moving towards each other. We are given their speeds, the observed frequency, and the speed of sound. We need to find the actual frequency of the source.
Step 2: Key Formula or Approach:
The general formula for the Doppler effect is: \[ f_{obs} = f_{actual} \left( \frac{v_{sound} \pm v_{observer}}{v_{sound} \mp v_{source}} \right) \] When the source and observer are approaching each other, the observed frequency increases. To get a factor greater than 1, we add the observer's velocity in the numerator and subtract the source's velocity in the denominator. \[ f_{obs} = f_{actual} \left( \frac{v_{sound} + v_{observer}}{v_{sound} - v_{source}} \right) \] We also need to convert the speeds from km/h to m/s by multiplying by \(\frac{5}{18}\).
Step 3: Detailed Explanation:
First, convert the velocities to m/s.
Velocity of source (car X), \(v_{source} = 36 \text{ km/h} = 36 \times \frac{5}{18} = 2 \times 5 = 10 \text{ m/s}\).
Velocity of observer (car Y), \(v_{observer} = 72 \text{ km/h} = 72 \times \frac{5}{18} = 4 \times 5 = 20 \text{ m/s}\).
Now, list the given values:
Observed frequency, \(f_{obs} = 1320\) Hz.
Velocity of sound, \(v_{sound} = 340\) m/s.
Substitute these values into the Doppler formula for approaching objects: \[ 1320 = f_{actual} \left( \frac{340 + 20}{340 - 10} \right) \] \[ 1320 = f_{actual} \left( \frac{360}{330} \right) \] Simplify the fraction: \[ \frac{360}{330} = \frac{36}{33} = \frac{12}{11} \] So, the equation becomes: \[ 1320 = f_{actual} \left( \frac{12}{11} \right) \] Now, solve for \(f_{actual}\): \[ f_{actual} = 1320 \times \frac{11}{12} \] \[ f_{actual} = \frac{1320}{12} \times 11 = 110 \times 11 = 1210 \text{ Hz} \] Step 4: Final Answer:
The actual frequency of the whistle sound is 1210 Hz.