Respuesta :
Answer
given,
mass of the block, m = 0.5 Kg
displacement, x = 30 cm = 0.3 m
Spring constant, k = 2 N/m
a) Angular frequency
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{2}{0.5}}[/tex]
[tex]\omega = 2\ rad/s[/tex]
b) Period of oscillation
[tex]T=\dfrac{2\pi}{\omega}[/tex]
[tex]T=\dfrac{2\pi}{2}[/tex]
T = 3.14 s
c) Position at t = 2
x = -A cos ω t
A = 0.3 ω = 2 rad/s
x = -0.3 cos (2 x 2)
x = 0.196 m
d) velocity at t= 2
v = A ω sin ω t
v = 0.3 x 2 x sin 4
v = -0.454 m/s
e) acceleration at t= 2
a = A ω² cos ω t
a = 0.3 x 2² cos 4
a = -0.784 m/s²
