Question

A student holds a tuning fork oscillating at 259 Hz. He walks toward a wall at a constant speed of 1.31 m/s. (a) What beat frequency does he observe between the tuning fork and its echo? Hz (b) How fast must he walk away from the wall to observe a beat frequency of 4.90 Hz? m/s

Answer 1

Answer (a).

STEP 1

USING DOPPLER FORMULA

Fd=F*(V+Vr)/(V-Vs) here Vr=Vs

Given that

F=259

V=346 (velocity of sound in air at room temperature)

As Vr=Vs =1.51

STEP 2

SUBSTITUTING THE VALUES

Fd= 256 * (346 + 1.51) / (346 - 1.51)

Fd= 88962.56/344.49

Fd= 258.244 Hz

STEP 3

FREQUENCY OF BEAT BETWEEN FORK AND ITS ECHO

Fbeat= |F-Fb|

Fbeat= 258.244 - 256

Fbeat= 2.244 Hz

Answer (b).

Fd= F-Fbeat

Fd= 256 - 4.9

Fd= 251.1 Hz

As we already calculated Fd now substitute each values in the Doppler formula

STEP 1

AGAIN USING DOPPLER FORMULA

Fd=F*(V-Vr)/(V+Vs) ,Vr = Vs

STEP 2

SUBSTITUTING THE VALUES

251.1 = 256 * (346 - Vr) / (346 + Vr)

251.1 = 88576 -256 Vr / 346 + Vr

251.1 * (346 + Vr) = 88576 -256 Vr

86880.6 + 251.1 Vr = 88576 -256 Vr

STEP 3

Arranging the values

251.1 Vr + 256 Vr = 88576 - 86880.6

507.1 Vr = 1695.4

Vr = 1695.4 / 507.1

Vr = 3.34 m/s