Question
Two sound sources are moving in opposite directions with velocities v1 and v2(v1>v2). Both are moving away from a stationary observer. The frequency of both the sources is 900 Hz. What is the value of v1–v2 so that the beat frequency observed by the observer is 6 Hz? Speed of sound v = 300 m/s. Given that v1 and v2 ≪ v.
Answer: Option B
:
B
f1=900(300300+v1)≅900(1+r1300)−1900−3v1
Similarly,
f2=900(300300+v2)=900−3v2f2−f1=6∴3(v1−v2)=6∴3(v1−v2)=6orv1−v2=2m/s
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:
B
f1=900(300300+v1)≅900(1+r1300)−1900−3v1
Similarly,
f2=900(300300+v2)=900−3v2f2−f1=6∴3(v1−v2)=6∴3(v1−v2)=6orv1−v2=2m/s
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