Astudent 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 (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

option a is correct i.e.

explanation:

in context to the membrane of nerve impulse, the outer membrane of neuron is semi permeable that is only sodium ions can pass through the membrane i.e. smaller ions can pass through the membrane while larger ions cannot pass through it.