Answer:
Part a)
[tex]A = 1.78 m[/tex]
Part b)
[tex]f = 2 rev/s[/tex]
Explanation:
Part A)
As we know that time period of the motion is given as
[tex]T = 2.68 s[/tex]
so we have
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{2.68}[/tex]
[tex]\omega = 2.34 rad/s[/tex]
now at the point of maximum amplitude the force equation when Normal force is about to zero is given as
[tex]mg = m\omega^2 A[/tex]
so we have
[tex]A = \frac{g}{\omega^2}[/tex]
[tex]A = \frac{9.81}{2.34^2}[/tex]
[tex]A = 1.78 m[/tex]
Part b)
Now if the amplitude of the SHM is 6.23 cm
and now at this amplitude if object will lose the contact then in that case again we have
[tex]mg = m\omega^2 A[/tex]
[tex]g = \omega^2 (0.0623)[/tex]
[tex]\omega = 12.5 rad/s[/tex]
so now we have
[tex]2\pi f = 12.5[/tex]
[tex]f = 2 rev/s[/tex]