Answer :
Answer:
T = 2.27 sec
Explanation:
the angular frequency of oscillations is given as
[tex]\omega = \sqrt{\frac{k}{m}[/tex]
k = spring constant, m = mass
[tex]=\sqrt {\frac{76}{6}[/tex]
[tex]\omega = 3.55 rad/sec[/tex]
time period is given as
[tex]T_0 = \frac{2\pi}{\omega}[/tex]
[tex]T_0 = \frac{2\pi}{3.55}[/tex]
T_0 = 1.76 sec
observer speed = 1.90*10^8 m/s
period of pendulum as calculated by observer is
[tex]T = \frac{T_0}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]
[tex]= \frac{1.76}{\sqrt{1- \frac{(1.90*10^8)^2}{(3*10^8)^2}}}[/tex]
T = 2.27 sec