Answer:
Explanation:
Given
mass of block [tex]m=1.6 kg[/tex]
Spring constant [tex]k=190 N/m[/tex]
initial displacement [tex] x_i=0.08 m[/tex]
Initial displacement is maximum displacement therefore
[tex] A=0.08 m[/tex]
General equation of motion of SHM is
[tex] x=A\cos \omega t[/tex]
where [tex]A=maximum\ amplitude[/tex]
[tex]\omega =Natural\ frequency\ of\ oscillation[/tex]
[tex]\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{190}{1.6}}=10.89[/tex]
differentiating x we get velocity which is given by
[tex]v=-A\omega \sin \omega t[/tex]
velocity at [tex]t=0.4 s[/tex]
[tex]v=-0.08\times 10.89\sin (10.89\times 0.4)[/tex]
[tex]v=0.817 m/s[/tex]
(b)time Period
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]T=2\pi \sqrt{\frac{1.6}{190}}[/tex]
[tex]T=0.576 s[/tex]
