Quiz 1

Class Vote:

As detailed in the class notes, if the index of refraction increases at a boundary, the wavelength of the EM radiation passing through decreases, while the frequency remains the same.  So the wavelength is inversely proportional to the index of refraction, n.  52% of the class got this right.
Quiz 2, first try:

Class Vote:

Congratulations to the 76 people in the class who got it right the first time!!  I was a little surprised that so many people, 77% of voters, chose A.  As a result, I spent some more of the class time than I normally would have describing the Doppler Effect, and what it is.  Keep in mind: the Doppler Effect is not an effect due to distance, in which as the distance gets smaller and smaller, the frequency gets higher and higher.  The Doppler Effect is due to speed.  If the speed is directly toward you, the frequency you hear is constant, and is higher than what you would hear if the object were at rest.  If the speed is directly away from you, the frequency you hear is constant, and is lower than the frequency you would hear if the object were at rest.


Quiz 3: Done on the Centre-screen, hand-written notes, stated:  Valerie is standing in the middle of the road, as a police car approaches her at a constant speed, v.  The siren on the police car emits a “rest frequency” of  f0.  Which statement is true?   

  1. The speed of the police car steadily rises as the police car gets closer and closer.
  2. The speed of the police car steadily decreases as the police car gets closer and closer.
  3. The speed of the police car does not change as the police car gets closer and closer.

Class Vote:

Good.  278 students agree that “constant speed” means the speed does not change.
Quiz 4 is the same as Quiz 2, except I asked it again:

Class Vote:

Much better!

Quiz 5:

Class Vote:

This was a very tricky quiz!  Congratulations to the 93 students who got this correct.  In fact, the most obvious answer, C, which is not quite correct, is very close, so please don’t feel bad if you chose C.  Personally, I am not able to solve this without consulting a calculator or doing some math.  Here is the solution:

In Valerie’s case, the observer is at rest while the source is approaching, so we use the first equation of 20.38 on page 636:

where fV is the frequency Valerie hears, fS is the rest frequency of the siren, v is the speed of the police car, and vsound is the speed of sound.  Let’s set x = v/vsound, where 0 < x < 1, and solve for the ratio of the frequency Valerie hears to the rest frequency:

In Daniel’s case, the source is at rest, and the observer is approaching it, so we use the first equation of 20.39 on page 637:

where fD is the frequency Daniel hears, and v is the speed of Daniel.  Let’s set x = v/vsound, and solve for the ratio of the frequency Daniel hears to the rest frequency:

So Valerie hears a frequency proportional to 1/(1−x), and Daniel hears a frequency proportional to 1+x.  Which is larger?  Well, you can easily check it on your calculator for several test values of x:

x

1/(1−x)

1+x

0.01

1.0101

1.0100

0.1

1.110

1.100

0.5

2.0

1.5

You can see that for x very small, they are almost exactly the same (C).  However, the frequency Daniel hears is always lower than the frequency Valerie hears (A).

 

If you don’t want to resort to a calculator, you could expand 1/(1−x) in a Taylor Series!  If you have a function, g(x), and x > 0 but x << 1, then you may approximate g(x) by the series:

where O(x3) are terms of order x3 or higher, which are small if x << 1. In our case, , g´(x)=(1−x)−2, and  g´´ (x)=2(1−x)−3, so the Taylor expansion becomes:

so, for very small x,

So the frequency Daniel hears is always lower than the frequency Valerie hears (A).