(a) In which of these scenarios do you hear a higher pitched sound? Explain. (i)

Question

(a) In which of these scenarios do you hear a higher pitched sound? Explain.

(i) A train with its horn sounding is stationary and you move directly towards it with a speed v0.

(ii) You are moving with speed v0/2 towards the same train and the train is sounding its horn as it moves directly towards you at speed v0/2.

(b) What is the difference in pitch between the above two scenarios if v0 = 30 m/s, the frequency of the train horn is 900 Hz and the air temperature is 0°C?

(c) What is the difference in pitch in (b) if (hypothetically) v0 is one-fourth the speed of sound?

Explanation / Answer

i) observed frequency is f1' = (V*f)/(V-Vs) = (V*f)/(V-(Vo/2)) = 2*v*f/(2v-Vo)

ii) f2' = (V+VL)*f/(V-Vs) = (V+(V0/2))*f/(V-(Vo/2)) = (2v+Vo)*f/(2v-Vo)

the frequency observed in ii) is greater than i)

so pitch is directly proportional to frequency

hence the pitch of the sound is higher in case ii)

b) V be the speed of sound at 0 deg C = 332 m/sec

then f1' = 2*v*f/(2v-Vo) = 2*332*900/((2*332)-30) = 942.6 Hz

f2' = (2v+Vo)*f/(2v-Vo) = ((2*332)+30)*900/((2*332)-30) = 985.17 Hz

difference is f2'-f1' = 985.17-942.6 = 42.57 Hz

C) if Vo = 332/4 = 83 Hz

f1' = 2*v*f/(2v-Vo) = 2*332*900/((2*332)-83) = 1028.57 Hz

f2' = (2v+Vo)*f/(2v-Vo) = ((2*332)+83)*900/((2*332)-83) = 1157.14 Hz

then

difference in pitch is 1157.14-1028.57 = 128.57 Hz

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