Given the equation of motion for the spring
X = 0.700cos(12.0t)
Mass of the body attached is
m = 0.56kg
(a) comparing the given equation to the general spring equation
X = A Cos(wt)
Where A is amplitude and w is frequency
Then, A = 0.7m
A = amplitude = 0.700m
(b) from the wave equation, we can deduced that,
The angular velocity is ω = 12rad/s
ω = √(k/m)
ω² = k/m
k = m×ω² = 0.56×12² = 80.6N/m
Spring constant k = 80.6N/m
(c) T = 2π/ω = 2π/12
T=π/6
At half period t = T/2
At t = T/2 = π/12
Since, X = 0.700cos(12.0t)
So, x = 0.700cos(12×π/12)
X = 0.700cos(π)
x = -0.7m
(d) at two-third of the period
t = ⅔ × 2π/12
t = π/9
X = 0.700cos(12.0t)
x = 0.700Cos(12×π/9)
x = 0.700Cos(4π/3)
x = 0.700 × -0.5
x = -0.35m
(e) to find t at x = -0.300m
–0.300 = 0.700cos(12t)
–0.300/0.700 = cos(12t)
cos(12t) = –0.429
12t = cos-¹(-0.429)
12t = 2.014
t = 2.01/12
t = 0.168s
