Answer
given,
mass of block = m = 2.3 Kg
spring constant = k = 150 N/m
speed = 2.3 m/s
a) we know,
[tex]\omega = \sqrt{\dfrac{k}{m}}[/tex]
[tex]\omega = \sqrt{\dfrac{150}{2.3}}[/tex]
ω = 8.08 rad/s
v = A ω
2.3 = A x 8.08
A = 0.285 m
b) maximum acceleration
a = A ω²
a = 0.285 x 8.08²
a = 18.58 m/s²
c) acceleration is minimum at mean position the minimum value of acceleration is zero.
d) time period
[tex]T = \dfrac{2\pi}{\omega}[/tex]
[tex]T = \dfrac{2\pi}{8.08}[/tex]
T = 0.778 s
t = 4.1 x 0.7778
t = 3.188 s