Question

A 6.00-kg object oscillates back and forth at the end of a spring whose spring constant is 76.0 N/m. An observer is traveling at a speed of 1.90 × 108 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?

Answer 1

T = 2.27 sec

Step-by-step explanation:

the angular frequency of oscillations is given as

`\omega = \sqrt{(k)/(m)`

k = spring constant, m = mass

`=\sqrt {(76)/(6)`

`\omega = 3.55 rad/sec`

time period is given as

`T_0 = (2\pi)/(\omega)`

`T_0 = (2\pi)/(3.55)`

T_0 = 1.76 sec

observer speed = 1.90*10^8 m/s

period of pendulum as calculated by observer is

`T = \frac{T_0}{\sqrt{1- (v^2)/(c^2)}}`

`= \frac{1.76}{\sqrt{1- ((1.90*10^8)^2)/((3*10^8)^2)}}`

T = 2.27 sec