An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 452 Hz when the train is approaching him and 442 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find

The velocity of the train is 3.84m/s

Explanation:

According to the Doppler effect, if the source is moving towards you then the apparent frequency of the sound emitted by the source is higher and if the source is moving away from you then the apparent frequency of the sound emitted by the source is smaller.

This is given by:

fo = V +-Vo/ V +-Vo × source

Where fo= observed frequency

V= velocity of sound

Vo= vo it of the observer

fsource= frequency the source

Given:

Observed frequency of the approaching train fo1= 452Hz

The observed frequency of train= fo2= 442Hz

Velocity of sound= 334m/s

Velocity of source=?

Train approaching the observer is given by:

fo1= V/(V - Vs)× source ...eq1

Train passes the student is given by:

fo2= V/(V - Vs)×source ...eq2

Divide eq1 by eq2

452/442 = (343+Vs)/(343 - Vs)

1.02 =(343+Vs)/(343 -Vs)

Cross multiply

1.02(343- Vs) = 343 + Vs

350.76 - 1.02Vs = 343 + Vs

Collecting like terms

350.76 -343= 1.02Vs+ Vs

7.76 = 2.02Vs

Vs= 7.76/2.02

Vs= 3.84m/s

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j