Question

Interactive Solution 28.5 illustrates one way to model this problem. A 6.52-kg object oscillates back and forth at the end of a spring whose spring constant is 51.6 N/m. An observer is traveling at a speed of 2.73 × 108 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?

Answer 1

We know that the angular frequency of spring mass system is given by
w = (K/M)^1/2
where w is angular frequency ,
K is spring constant and
M is mass of the object
w = (51.6/6.52)^1/2 = 2.813 rad/s
We know that
w = 2 / T
where T is the time period
T = 2 /w = 2 / 2.813 = 2.233 s
Now the relativistic time period is given by
T' = T /[1 -(v/C)²]^1/2
Where v is speed of observer = 2.73*10⁸ m/s
C is speed of light = 3*10⁸
T' = 2.233 /[1 - (2.73*10⁸/3*10⁸)²]^1/2
T' = 5.386 s