Question

A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at a speed of 2.80 *10^8 m/s relative to the fixed end ofthe spring. What does this observer measure for the period ofoscillation.
The answer in the back of the book is 5.57 seconds, but I keepgetting 30 seconds. Can someone help tell me what I am doing wronghere is my work:
t= period observer sees
to=real period
to=2π*√(m/k)=2π*√(5.00kg/49.3N/m)=2.00sec
t=to/(1-√(v^2/c^2)), where v=2.80*10 ^8 m/s andc=3.0*10^8 m/s
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))=30seconds

Answer 1

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec