Answer:
(a) the amplitude of the oscillations is 0.52 m
(b) position of the block at t = 0 s is - 0.06 m
(c) the velocity of the block at t = 0 s is 3.96 m/s
Explanation:
given information:
m = 4.10 kg
k = 240 N/m
t = 1.70 s
x = 0.165 m
v = 3.780 m/s
(a) the amplitude of the oscillations
x(t) = A cos (ωt+φ)
v(t) = dx(t)/dt
= - ω A sin (ωt+φ)
ω = [tex]\sqrt{\frac{k}{m} }[/tex]
= [tex]\sqrt{\frac{240}{4.10} }[/tex]
= 7.65 rad/s
v(t)/x(t) = - ω A sin (ωt+φ)/A cos (ωt+φ)
v(t)/x(t) = - ω sin (ωt+φ)/cos (ωt+φ)
v(t)/x(t) = - ω tan (ωt+φ)
(ωt+φ) = [tex]tan^{-1}[/tex] (-v/ωx)
= [tex]tan^{-1}[/tex] (-3.780/(7.65)(0.165))
= - 1.25
(ωt+φ) = - 1.25
φ = - 1.25 - ωt
= - 1.25 - (7.65 x 1.70)
= - 14.26
x(t) = A cos (ωt+φ)
A = x(t) / cos (ωt+φ)
= 0.165/cos(-1.25)
= 0.52 m
(b) position, at t=0
x(t) = A cos (ωt+φ)
x(0) = A cos (ω(0)+φ)
x(0) = A cos (φ)
= 0.52 cos (-14.26)
= - 0.06 m
(c) the velocity, at t=0
v(t) = - ω A sin (ωt+φ)
v(0) = - ω A sin (ω(0)+φ)
= - ω A sin (φ)
= - (7.65)(0.52) sin (-14.26)
= 3.96 m/s
