What is gamma for an object with speed B = 0.1?

  • Thread starter Thread starter dranger35
  • Start date Start date
  • Tags Tags
    Gamma Speed
Click For Summary
Gamma for an object with speed B = 0.1 is calculated using the formula gamma = 1/sqrt(1-B^2), resulting in a value of approximately 1.005. For the second problem, the velocity addition formula in special relativity shows that the speed of a car moving at 0.05c relative to an observer at rest is approximately 0.1c when calculated correctly. The third problem involves determining the area of a coil by multiplying its length by the number of loops, which is 0.1 m^2, and using this to calculate the minimum angular frequency (omega) that can be measured, yielding approximately 4.8x10^15 radians/second. Understanding these concepts is crucial for applying special relativity principles effectively.
dranger35
Messages
14
Reaction score
0
Hey I was given 3 problems for homework, I just need help on some and the others I want to make sure i did it right.

Problem 1: What is gamma for an object with speed B = 0.1?

Problem 2: You are at rest, and see me as moving with a speed of
B = 0.05. I see a car moving with a speed of B = 0.05. How fast do
you measure the car to be moving? In particular, how much
different is the car's speed from B = 0.1?

Problem 3: Assume that you can measure the phase difference to 1 degree.
What is the minimum w(omega) that you can measure? Assume light with
a wave length of 1000 nm, and a 0.1 m2 coil with 1000 loops.


for 1 I know gamma=1/sq. root(1-B^2). Now is that all you do, because it just seems to easy.

for 2 I 'm not sure how to start it

for 3 I know difference in phase= 8(pi)(Area)(3 of loops)(omega) /c(wavelength) , but how do i find the area?
 
Physics news on Phys.org
Problem 1: Dead simple. Just look at the definitions of gamma and beta. The object's velocity is 0.1c, or one-tenth the speed of light. How do I know? Because \beta is defined as the ratio of an object's speed to the speed of light: v/c = 0.1c/c = 0.1. Now tell me? What is the factor gamma, and why is it important? (This is an integral part of special relativity here. If you can't do this problem right away without even thinking about it, then I'd recommend reviewing the def'ns of gamma and beta). *EDIT* Oh...you did do the problem without thinking about it too much. Good! :smile: Well then, to answer your question, yes, the problem really is that easy...a matter of simple substitution. But I want to re-emphasize that you make sure you know what these factors mean.

Problem 2: Start like this. You're "at rest". (Why did I put this in quotes? :wink: ) Call your inertial frame of reference S. You see me moving at beta = 0.05. So, call my inertial frame of reference S'. I (and my frame) am therefore moving at v' = 0.05c relative to you. Now you see another car, also moving at 0.05c (call this velocity: u). So u relative to you is 0.05c. What is u' (relative to me?) i.e, if the velocity u is 0.05c in the S frame, how do I calculate this velocity as measured in the S' frame? (u').

Problem 3 seems to be worded too vaguely for me, at least, to figure out what's going on.
 


For problem 1, yes, you are correct. To find gamma for an object with speed B=0.1, you would simply plug in 0.1 for B in the equation gamma=1/sq. root(1-B^2). So in this case, gamma would be approximately 1.005.

For problem 2, we can use the formula for velocity addition in special relativity: v'=(v+u)/(1+vu/c^2), where v is the velocity of the car as measured by you, u is the velocity of the car as measured by me, and c is the speed of light. Since both of our velocities are 0.05, we can plug these values in to get v'=(0.05+0.05)/(1+0.05*0.05/1)=0.099975. So the speed of the car as measured by you would be approximately 0.1, which is the same as B=0.1.

For problem 3, the area of the coil can be found by multiplying the length of the coil (0.1m) by the number of loops (1000), giving us an area of 0.1 m^2. Plugging this into the formula for difference in phase, we get difference in phase= 8(pi)(0.1 m^2)(1000 loops)(omega)/c(1000 nm). To find the minimum w(omega) that can be measured, we can set the difference in phase to 1 degree (since that is the smallest unit we can measure) and solve for w. This would give us a minimum w(omega) of approximately 4.8x10^15 radians/second.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
943
  • · Replies 44 ·
2
Replies
44
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
9
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
Replies
7
Views
1K