Answer:
a) m = 0.626 kg
, b) T = 2.09 s
, c) a = 1.0544 m / s²
Explanation:
In a spring mass system the equation of motion is
x = A cos (wt + Ф)
with w = √(k / m)
a) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф) (1)
give us that the speed is
v = 26.1 m / s
for the point
x = a / 2
the range of motion is a = 11.0 cm
x = 11.0 / 2
x = 5.5 cm
Let's find the time it takes to get to this distance
wt + Ф = cos⁻¹ (x / A)
wt + Ф = cos 0.5
wt + Ф = 0.877
In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f
Ф = 0
wt = 0.877
t = 0.877 / w
we substitute in equation 1
26.1 = -11.0 w sin (w 0.877 / w)
w = 26.1 / (11 sin 0.877))
w = 3.096 rad / s
from the angular velocity equation
w² = k / m
m = k / w²
m = 6 / 3,096²
m = 0.626 kg
b) angular velocity and frequency are related
w = 2π f
frequency and period are related
f = 1 / T
we substitute
w = 2π / T
T = 2π / w
T = 2π / 3,096
T = 2.09 s
c) maximum acceleration
the acceleration of defined by
a = dv / dt
a = - Aw² cos (wt)
the acceleration is maximum when the cosine is ±1
a = A w²
a = 11 3,096²
a = 105.44 cm / s²
we reduce to m / s
a = 1.0544 m / s²
