A5.00 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 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/ 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

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

a.metals can become magnetic if they have many spinning electrons that create magnetic fields.d.only some metals are attracted by magnets; these metals are called ferromagnetic.

d.wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.