Answer:
(a) 0.993 s
(b) 14.0 N/m
(c) -3.02 m/s
(d) -6.01 m/s²
Explanation:
(a) The block's position can be modeled as a cosine wave:
x(t) = A cos(ωt)
where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.
At t = 0.200 s, x(t) = 15.0 cm.
15.0 cm = 50.0 cm cos((0.200 s) ω)
0.3 = cos((0.2 s) ω)
1.266 rad = (0.2 s) ω
ω = 6.33 rad/s
The period is:
T = (2π rad) (1 s / 6.33 rad)
T = 0.993
(b) For a spring-mass system, ω = √(k/m). The mass of the block is 0.350 kg, so:
ω = √(k/m)
6.33 rad/s = √(k / 0.350 kg)
6.33 rad/s = √(k / 0.350 kg)
40.1 rad/s² = k / 0.350 kg
k = 14.0 N/m
(c) Energy is conserved:
EE₀ = EE + KE
½ kx₀² = ½ kx² + ½ mv²
kx₀² = kx² + mv²
(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²
v = -3.02 m/s
Alternatively, we can take the derivative of our position equation:
v(t) = -Aω sin(ωt)
v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))
v = -3.02 m/s
(d) Sum of forces on the block:
∑F = ma
-kx = ma
a = -kx / m
a = -(14.0 N/m) (0.15 m) / (0.350 kg)
a = -6.01 m/s²
Alternatively, we can take the derivative of our velocity equation:
a(t) = -Aω² cos(ωt)
a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))
a = -6.01 m/s²
