AY2 Homework Set 3 SOLUTIONS


Answer all of the five questions.  Show your work for the problems
that need math.
Useful formulas:

     E =h \nu  (\nu = frequency  & h= 6.6 x10^ -34 Joules/sec )

     c = \nu \lambda       c =  3 x 10^8 m/s

     t (moving) = t (rest) * squareroot  [1- (v/c)^2 ]


1. If you watch your friend moving by, you'll say that her time is
running slow, her length is contracted, and her mass is grater than
her rest mass. 

How will she perceive her own time, length, and mass? Why? 

She will not see a change in her own time, length and mass, because
she is at rest in the moving frame of reference.  

How will she perceive your time, length, and mass?

Now, she will perceive us contracting in length, growing in mass and
our time will seem to slow down from her point of view; she can say
that she is at rest and we are moving.   So she sees us as we see her
and vice-versa.


2. The famous Martian National Public Station Super Estrella broadcast
at 107.1 FM.  

If your were to find it on the radio dial, what would
its frequency be in MHz (megahertz) given that the radio signal has a
wavelength of  2.8 meters?  

 1 Hz ( 1/sec or sec-1)

 \nu = c / \lmbda =3 x 10^8 m/s / 2.8 m =107.1 x 10^6 sec-1=107.1 Mhz

If Super Estrella also broadcast at  1602
AM (1602 Kilohertz)  what is its wavelength of the radio signal? 

 \lambda = c / \nu = 3 x 10^8 m/s / 1602 *10^3 Hz =   187 m

For which band  ( FM or AM ) does the radio wave have more energy?

The FM   band has more energy since the \lambda is shorter and \nu is 
larger than the AM band.


3. A "clever" student after learning about the theory of relativity
decides to apply his knowledge in order to prolong his life.  He
decides to spend the rest of his life, let's say 55 years, in a
bicycle traveling around the neighborhood at 20 miles per hour.  How
much time will pass on earth, in other words, outside the bicycle in
the "non moving reference frame"?

In problem 3 & 4,
     t_rest = t_Earth
     t_moving = time in reference frame of moving student
     t (moving) = t (rest) * squareroot  [1- (v/c)^2 ]
     t_rest = t_moving / squareroot  [1- (v/c)^2 ]

All units cancel out  and we are only left with " the units of years "

  c = 3 x 10^8 m/s  =  186000 miles / hr
  t_rest = t_moving / squareroot  [1- (v/c)^2 ]
  t_earth = 55 yr /  squareroot  [1- (20 miles / hr / 186,000 miles / hr)^2 ]

You see, what happens is that since we are going so slow relative to
the speed of light, the operation in the botton (denominator; 
squareroot  [1- (v/c)^2 ]) is almost 1   (1.0000000000000001).  So 55 devided by that number just
gives us 55 again, telling us that at that slow speed there will be NO
change   in time.

4. Another maniac-waco student decides to spend time cruising around
the local solar neighborhood at a speed of 0.90 c  (90% the speed of
light).  How much time will go by on earth if in her spacecraft 60
years pass in the student's clock?

Before you begin the math, think about what kind of answer you
expect.  Do you expect the time that passes on Earth to be greater than or less
than the time on the student's clock?  It should be greater than that
clock because from the Earth's (non-moving) perspective, the student's
moving clock will appear to be going slower.

  t_rest = t_moving / squareroot  [1- (v/c)^2 ]
  t_earth = 60 yr /  squareroot  [1- (1 - (0.9 c/ c)^2)] 

====> t(Earth) = 138 years  

In other words, she is going so fast that on earth passed 138 years while
on her spaceship only 60 years.

5. On a realistic trip to the stars, we could not suddenly jump close
to the speed of light (like in Star Wars) without being killed by the
huge force of such acceleration (too many g-forces to handle).  Thus,
a more realistic trip would have us accelerate at a comfortable rate,
such as 1 g, until we are half way to our destination and then
decelerate at the same rate until we reach our destination.  Explain
why we would be comfortable with this acceleration. 

This problem has to do with the general theory of relativity.  The
principle of equivalence is explained in your text p. 407-408.  It
states that the effects of gravity are exactly equivalent to the
effects of acceleration.  
g=9.8 m/s^2. "g" is the acceleration on the surface of the earth due
to the force of gravity. So acceleration in a rocket ship at "1 g"
will be very comfortable because it is the acceleration of gravity for
objects when they fall on earth.  In other words, we would experience
the same type forces as we walk and live daily in earth. 

By our own
reckoning, would we notice anything unusual about lengths, masses, or
the passage of time on our spaceship? Why or why not?
For the second part just refer to question one.