Answer:
Explanation:
Given that,
Mass attached to spring
M = 0.52kg
Force constant K = 8N/m
Amplitude A = 11.6 cm
a. Maximum speed?
Angular velocity is calculated using
w = √k/m
w = √8/0.52
w = √15.385
w = 3.922rad/s
Then, the relation ship between angular velocity and linear velocity is given as
v = - w•A
v = - 3.922 × 11.6
v = - 45.5 cm/s
Then, the maximum velocity is
vmax = |v|= 45.5cm/s
b. Acceleration a?
Acceleration can be determine using the formula
a = -w²• A
a = -3.922² × 11.6
a = -178.46 cm/s²
Magnitude of the acceleration is 178.46cm/s²
c. Speed when the object is at 9.6cm from equilibrium position?
Generally,
The position of the object at equilibrium is
x(t) = A•Cos(wt)
x(t) = 11.6 Cos (3.922t)
Then, when x(t) = 9.6cm
9.6 = 11.6 Cos(3.92t)
Cos(3.922t) = 9.6/11.6
Cos(3.922t) = 0.8276
3.922t = ArcCos(0.8276)
Note: the angle is in radiant
3.922t = 0.596
t = 0.596/3.922
t = 0.152 second
Then, v(t) at that time is
v(t) = x'(t) = -11.6×3.92Sin(3.922t)
v(t) = -45.5Sin(3.922t)
Now, when t =0.152
v(t) = -45.5 Sin(3.922×0.152)
v(t) = -45.5Sin(0.596)
v(t) = -25.5 cm/s
Then, it's magnitude is 25.5cm/s
d. Acceleration at same position
t = 0.152s
a(t) = v'(t) = - 45.5×3.922Cos(3.922t)
a(t) = -178.46Cos(3.92t)
a(t) = -178.46 Cos(3.92×0.152)
a(t) = -178.46 Cos(0.596)
a(t) = -147.68 cm/s²
Magnitude of the acceleration is 147.68 cm/s²
