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A toy car, blowing its horn, is moving with a steady speed of 5 m/s, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m/s, the frequency of the horn of the toy car is close to:
a) 680 Hz
b) 510 Hz
c) 340 Hz
d) 170 Hz

Answer
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Hint: Let us assume the frequency when the train is approaching is f1 and when the train is leaving is f2. Get the value of f1 and f2 in terms of f. Now, as we know that beat frequency is given by: B= f1f2, find the frequency.

Formula used:
1. Frequency when source is approaching: f=f(vvvs)
2. Frequency when source is leaving: f=f(vv+vs)
where f is the frequency required, f is the frequency given, v is the velocity of observer and vs is the velocity of sound.
3. B= f1f2, where B is the beat frequency and ( f1f2) is the change in frequency.

Complete step by step answer:
We have:
Frequency = f
Velocity of Source (train) vs=5ms1
Velocity of sound v=340ms1
Beats: B= 5s1
So, by using the formula:
Frequency when source is approaching: f=f(vvvs)
Frequency when source is leaving: f=f(vv+vs)
We get:
f1=f(3403405)......(1)
f2=f(340340+5)......(2)
As we know that:
B= f1f2
So, we have:
5=f(3403405)f(340340+5)5=340f(13351345)5=340f(10335×345)f=5×335×345340×10=169.9Hz

So, the correct answer is “Option D”.

Note:
The formula of Doppler Effect related the frequency of the sound of an object with its velocity. Doppler Effect states that the change in wave frequency during the relative motion between a wave source and its observer. It was discovered by Christian Johann Doppler. Beats can be calculated by subtracting the initial frequency with the frequency observed by the observer.
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