Devon speeds past Tamara in his space shuttle. In Tamara's frame of reference, the shuttle moves with a speed of 0.88c. Devon's radio is blaring out music with a beat that occurs every 0.50 seconds. How often does Tamara hear the beats?
I'm getting 0.24 s but it's saying that's wrong...
0.50=t./(the root of: 1-(0.88c^2/c^2))
0.50=t./0.474974
t.=0.2375
Thanks for your time :)
remember the twins paradox?
one stays on Earth, the other goes on a high speed (close to c) trip
the one on Earth grows old, but when the traveler returns, he is still young
time passes slower in the high speed reference frame
so the beats will be wider spaced for Tamara
moving time = fixed time * dilation factor
0.50 = t * 0.474974
not sure why you had division, not multiplication
You're right that does make sense, I got confused & mixed up t & t. if I switch them I get 1.1s.
Thank you :) I appreciate your help
To answer this question, we need to consider the concept of time dilation in special relativity. Time dilation states that time appears to pass more slowly for objects in motion relative to an observer at rest.
We are given that in Tamara's frame of reference, Devon's space shuttle is moving at a speed of 0.88c, which means it is traveling at 0.88 times the speed of light (c).
The time dilation equation for time measured by an observer at rest is given by:
t' = t / sqrt(1 - (v^2 / c^2))
Where:
t' is the time measured by the observer at rest (Tamara)
t is the time measured by the moving object (Devon)
v is the relative velocity (0.88c in this case)
c is the speed of light in a vacuum
Let's calculate the value for t' using this equation:
t' = 0.50 / sqrt(1 - (0.88^2))
t' = 0.50 / sqrt(1 - 0.7744)
t' = 0.50 / sqrt(0.2256)
t' = 0.50 / 0.474974
t' ≈ 1.052 s
Therefore, Tamara would hear the beats occurring every 1.052 seconds, not 0.24 seconds as initially calculated.