Respuesta :
Answer:
(A) 89.442 rad/sec
(B) 0.070 sec
(C) [tex]C=25\times 10^{-6}F[/tex]
Explanation:
We have given mass of the body m = 0.50 kg
It is given that spring is extended 2 mm from its equilibrium position
So x = 2 mm = 0.002 m
Resistive force is given F = 8 N
We know that force is equal to F = Kx, here K is spring constant and x is elongation in spring
So [tex]8=K\times 0.002[/tex]
K = 4000 N/m
(A) Angular frequency is equal to [tex]\omega =\sqrt{\frac{K}{m}}[/tex]
So angular frequency [tex]\omega =\sqrt{\frac{4000}{0.50}}=89.442rad/sec[/tex]
(b) Time period of oscillation is equal to [tex]T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{89.442}=0.070sec[/tex]
So time period will be 0.070 sec
(c) Inductance is given L = 5 H
Angular frequency is equal to [tex]\omega =\frac{1}{\sqrt{LC}}[/tex]
[tex]89.442 =\frac{1}{\sqrt{5C}}[/tex]
Squaring both side
[tex]8000=\frac{1}{5C}[/tex]
[tex]C=25\times 10^{-6}F[/tex]
