A 3kg block is attached to an ideal spring with a
force constant k = 200N/m. The block is given
an initial velocity in the positive direction of
magnitude u = 12 m/s and no initial
displacement (x o = 0). Find (10 marks)
a. The amplitude and
b. The phase angle
c. Write an equation for the position as a
function of time
depends on your definitions but if
x = A sin (wt-p)
u = A w cos (wt-p)
a = -A w^2 sin(wt-p) = -w^2 x
if x = 0 at t = 0
then
x = A sin (-p) = 0 and a solution is p = 0
x = A sin (wt)
u = A w cos (wt)
a = -Aw^2 sin wt = -A w^2
now u = 12 when t = 0
A w cos(0) = 12
so
A w = 12
now k = 200 N/m
F = m a = -m (Aw^2)sin w t = - k x = -k (A sin (wt))
so
w^2 = k/m ( which you probably knew :)
w^2 = 200 /3
w = 10 sqrt(2/3)
A = 12/w = 0.1 sqrt(1.5)
x = 0.1 sqrt 1.5 sin ( 10 sqrt(2/3) t)
To find the amplitude, phase angle, and the equation for the position as a function of time, we can use the concepts of simple harmonic motion (SHM) and equations derived from it.
a. The amplitude of an oscillating motion is the maximum displacement from the equilibrium position. In this case, since the block is attached to an ideal spring, the amplitude can be determined by considering the energy of the system. The total mechanical energy (E) of the system is given by the sum of the kinetic energy (K) and the potential energy (U) due to the spring:
E = K + U
Initially, the block is given an initial velocity, so K = 1/2 mu^2, where m is the mass of the block and u is its initial velocity. Since the block is attached to an ideal spring, the potential energy is given by U = 1/2 kx^2, where k is the force constant of the spring and x is the displacement of the block from the equilibrium position.
At the maximum amplitude, the block momentarily comes to rest, so its kinetic energy is zero. At this point, all the mechanical energy is stored in the potential energy of the spring. Therefore, we have:
E = U_max
1/2 mu^2 = 1/2 kx_max^2
Rearranging the equation, we can solve for x_max, which represents the amplitude:
x_max = sqrt((mu^2) / k)
x_max = sqrt((3kg)(12 m/s)^2 / 200 N/m)
x_max = sqrt(3.6 m^2)
x_max ≈ 1.90 m
Therefore, the amplitude of the motion is approximately 1.90 m.
b. The phase angle (φ) represents the initial displacement of the block from the equilibrium position when the motion starts. In this case, the initial displacement (x_0) is given as 0 (no initial displacement). Thus, the phase angle is zero.
c. To write the equation for the position as a function of time, we can use the equation:
x(t) = x_max * cos(ωt + φ)
where x(t) is the position of the block at time t, ω is the angular frequency of the oscillation, and φ is the phase angle.
The angular frequency (ω) can be calculated using the equation:
ω = sqrt(k / m)
In this case, when substituting the given values:
ω = sqrt(200 N/m / 3 kg)
ω ≈ 8.16 rad/s
Since the phase angle (φ) is zero, the equation for the position as a function of time becomes:
x(t) = 1.90 m * cos(8.16 rad/s * t)
This equation represents the position of the block as it undergoes simple harmonic motion with the given initial conditions.