Question
I KNOW THE ANSWERS BUT I NEED THE WORK AND EXPLANATION FOR 5 STARS!
The position of a simple harmonic oscillator is given by x(t) = (2.4m)cos . What is the period of the oscillator? 10 s A bug is at the center of the turntable. With the turntable is rotating at an angular frequency of 4 rad/s, the bug decides to run to the outer edge of the turntable, resulting in a new angular frequency of 3.5 rad/s. If the turntable has a moment of inertia of 0.006 kg.m2 and a radius of 0.68 m, what is the mass of the bug? 1.9 times 10-3 kg
Explanation / Answer
8)For x=Asin(wt)
T=2*pi/w
so Here T=2*pi/(pi/5)=10s
9)Initital ang. momentum=final ang.momentum
So 0.006*4=(0.006+mr^2)*3.5
So m=1.9*10^-3 Kg