Answer:
Explanation:
given,
mass of block = 1.2 kg
spring constant (k) = 480 N/m
a) Frequency
[tex]f = \dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]
[tex]f = \dfrac{1}{2\pi }\sqrt{\dfrac{480}{1.2}}[/tex]
f = 3.183 Hz
b) The position function
x = A cos (ωt + ∅)
at t = 0 , x = 0
0 = A cos ∅
cos ∅ = 0
∅ = π/2
velocity at t= 0 s
[tex]v = \dfrac{dx}{dt}[/tex]
v = Aωsin (ωt + ∅)
[tex]v = A\omega sin ( + \dfrac{\pi }{2})[/tex]
[tex]5.2 = A\sqrt{\dfrac{480}{1.2}}[/tex]
A = 0.26 m
c) [tex]x = 0.26 cos (\sqrt{\dfrac{480}{1.2}}t + \dfrac{\pi }{2})[/tex]
[tex]x = 0.26 cos( 20 t +\dfrac{\pi }{2})[/tex]
