A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?


Answer 1

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 ((v_0)/(v_0-v))


f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 ((v_0)/(v_0-v))

300 = f_0 ((343)/(343-v))

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 ((v_0)/(v_0-v/2))

290 = (343)((v_0)/(343-v/2))

290*343-145v = 343f_0

Dividing the two expression we have,

((300*343) - 300v)/(290*343-145v) = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

Answer 2

Final answer:

The speed of the train can be determined using the Doppler effect formula.


The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))


  • f' is the observed frequency
  • f is the source frequency
  • v is the speed of sound
  • v_o is the speed of the observer (here it is the train)
  • v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

Learn more about Doppler effect here:



Related Questions

The planet earth orbits around the sun and also spins around its own axis. 33% part (a) calculate the angular momentum of the earth in its orbit around the sun in kg • m2/s
A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?
The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.
What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?
After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pulls his arms towards his body to reduce his MOMENT OF INTERTIA by 1/2, what is his ANGULAR VELOCITY w? c) How much does his ROTATIONAL KINETIC ENERGY change? That is, if the initial Kinetic Energy is (KE)initial, what is the final KE? d) Where did that ENERGY come from, or go to?




a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final =  1/2 I₂ ω₂²

= 1/2 (I₁ / 2)  (2ω)²

=  I₁ ω²

c )

Change in rotational kinetic energy

=  I₁ ω² -  1/2 I₁ ω²

=  +  1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

A glass plate 2.95 mmmm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 600 nmnm (in vacuum) and a screen. The distance from source to screen is 1.25 cm. How many wavelengths are there between the source and the screen?






Thickness of the glass plate, x=2.95* 10^(-3)\ m

refractive index of the glass plate, n=1.6

wavelength of light source in vacuum, \lambda=600* 10^(-9)\ m

distance between the source and the screen, d=1.25\ m

Distance travelled by the light from source to screen in vacuum:



d_v=1.24705\ m

So the no. of wavelengths in the vacuum:


N=(1.24705)/(6* 10^(-7))

N\approx2.0784* 10^(6)  .......................(1)

Now we find the wavelength of the light wave in the glass:



\lambda'= wavelength of light in the medium of glass.

1.6=(600* 10^(-9))/(\lambda')

\lambda'=375* 10^(-9)\ m=375\ nm

Now the no. of wavelengths in the glass:


N'=(2.95* 10^(-3))/(375* 10^(-9))

N'=7.8667* 10^(3) ............................(2)

From (1) & (2):

  • total no. of wavelengths are there between the source and the screen:



4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]


The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t


θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration =  4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec

Hence the angular velocity will be 22.41 rad/s.

To learn more about angularvelocity refer to the link



w_f= 22.41rad/s


First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:


where w_f is the final angular velocity, w_i the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:


Finally, solving for w_f:

w_f= 22.41rad/s

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?


We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations          (1 minute  = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

6. As distance increases, gravitational force *
(10 Points)


It decreasessssssssss

The measurement of an electron's energy requires a time interval of 1.2×10^−8 s . What is the smallest possible uncertainty in the electron's energy?



1.05* 10^(-26)J


The uncertainty in energy is given by \Delta E=(h)/(2\pi \Delta t)

here h is plank's constant which value is 6.67* 10^(-34) and \Delta t is the time interval which is given as 1.2* 10^(-8)sec

So using all the parameters the smallest possible uncertainty in electrons energy is =(6.67* 10^(-34))/(2* \pi * 1.2* 10^(-8))=1.05* 10^(-26)J