GREPhysics.NET
GR | # Login | Register
   
  GR8677 #12
Problem
GREPhysics.NET Official Solution    Alternate Solutions
Verbatim question for GR8677 #12
Wave Phenomena}Doppler Effect

For a source that's coming towards you, its frequency will obviously be higher than its original source frequency. Choices A and B are out. To know for sure whether the choice is C, D, or E, one can apply the Doppler Effect.

The Doppler Effect equation can easily be derived even if one forgets it. A source f_0 approaching at v_0 would have \Delta \lambda, the distance between waves, decreasing. v=\lambda f\Rightarrow \Delta \lambda=\frac{v_{sound}-v_{0}}{f_{0}}. The \Delta \lambda wave travels at the speed of sound, thus the frequency you receive is: f=\frac{v_{sound}}{\Delta \lambda}=10f_0. Choice E.

See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
flyboy621
2010-11-09 21:01:01
I find it easier to look at the time between successive wave crests. Let's say if the source were at rest, the time between crests would be T. Then if the source is moving toward the observer at .9c_s, the time between crests is T/10. Therefore the frequency is 10 times higher, as in (E).Alternate Solution - Unverified
kicksp
2007-10-29 10:11:11
Quick hand-waving arguments save us time here. Note that the observed frequency f goes to infinity as v\to v_0. A simple functional form for f (which we could construct immediately even if we forgot the formula) is

f\prop \frac{1}{(1-v/v_0)},

yielding (E).

Go maroons!
Alternate Solution - Unverified
Comments
flyboy621
2010-11-09 21:01:01
I find it easier to look at the time between successive wave crests. Let's say if the source were at rest, the time between crests would be T. Then if the source is moving toward the observer at .9c_s, the time between crests is T/10. Therefore the frequency is 10 times higher, as in (E).Alternate Solution - Unverified
student2008
2008-10-11 12:48:31
...the frequency of the wave is \omega_0 = \omega - k \cdot v_s....NEC
student2008
2008-10-11 12:44:23
The observable phase of the wave front at some point is \phi_s = \omega \cdot t - k \cdot x, where \omega and k are the circular frequency and the wave vector measured in the observational reference frame. The source is moving with the constant velocity v_s, so x = v_s  \cdot t. Thus, \phi_s = (\omega - k \cdot v_s) t. It means that in the reference frame where the source is at rest the frequency of the wave is \omega_0 = \omega - k \cdot c. Since k = {\omega \over c} , we get eventually:  \omega = \frac{\omega_0}{1 - \frac{v_s}c}.NEC
ssp
2008-09-05 02:46:12
Bit more handwaving...

So we take the fact that we know the frequency has to increase from above. Now we look at the speed. The shift has to large because the speed is large (speed and frequency are still proportional) and you got your answer, (E).

Little trick from GR
NEC
kicksp
2007-10-29 10:11:11
Quick hand-waving arguments save us time here. Note that the observed frequency f goes to infinity as v\to v_0. A simple functional form for f (which we could construct immediately even if we forgot the formula) is

f\prop \frac{1}{(1-v/v_0)},

yielding (E).

Go maroons!
vikhuan
2009-06-08 02:04:49
tyjyyty
Alternate Solution - Unverified

Post A Comment!
Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...