Answer:
a) T = 1.26 s , b) v_max = 0.14 m / s , a_max = 0.7 m / s²
c) x = 0.028 cos (5 t) , v = - 0.14 sin 5t, a = - 0.7 cos 5t
Explanation:
This is a simple harmonic motion exercise that is described by the equation
x = A cos (wt +Ф)
with
w = √ (k / m)
let's apply this expression to our case
a) Angular velocity is related to frequency
w = 2π f
frequency and period are related
f = 1 / T
we substitute
2π / T = √ (k / m)
T = 2π √(m / k)
let's calculate
T = 2π √(1/25)
T = 1.26 s
In the expression for the period, the amplitude does not appear, therefore there is no dependence, as long as Hooke's law is fulfilled, which is correct for small amplitudes.
b) in the initial equation we have the position as a function of time, let's use the definition of speed and acceleration
v = dx / dt
v = - A w sin (wt + Ф)
the speed is maximum when the sine is -1
v_max = A w
w = √ (k / m)
w = √ 25/1
w = 5 rad / s
the amplitude of the movement is equal to the maximum compression of the spring
A = 2.8 cm = 0.028 m
we substitute
v_max = 0.028 5
v_max = 0.14 m / s
acceleration
a = dv / dt
a = - A w² cos (wt + Ф)
the acceleration is maximum when the cosine is -1
a_max = A w²
let's calculate
a_max = 0.028 5²
a_max = 0.7 m / s²
c) let's start by finding the phase constant
v = -A w cos (wt + Ф)
at t = 0 they indicate that the system has v = 0
0 = -A w sin (0 + Ф)
Ф = sin⁻¹ 0
Ф = 0
we write the equation
x = 0.028 cos (5 t)
v = - A w sin (wt + Ф)
v = - 0.028 5 sin (5t + 0)
v = - 0.14 sin 5t
acceleration
a = - A w² cos (wt + Ф)
a = - 0.028 5 2 cos (5 t + 0)
a = - 0.7 cos 5t
