Final Answer:
The initial displacement of the weight from equilibrium at \(t = 0\) is 2 meters upwards. This measurement considers the position of the weight at t = 0 relative to the equilibrium position.
Step-by-step explanation:
The equation for the displacement of the weight attached to the spring can be determined using the equation of motion for simple harmonic motion:
[y(t) = A. cosω t + frac{
}{k} + frac{
}{ω}. sin(ω t)]
Where:
y(t) = displacement at time t
A = amplitude of the oscillation
ω = angular frequency
= amplitude of the external force
k = spring constant
= initial velocity
Given A = 2 meters (initial displacement), k = 72 kg/m, v_0 = 1 m/s, and F(t) = -450 sin(3t), we can find the angular frequency ω using ω= √(k/m) where m is the mass.
The mass m is 18 kg. Thus, ω = sqrt{frac{72}{18}} = sqrt{4} = 2.
Plugging the values into the displacement equation at t = 0:
[y(0) = 2.cos(0) + frac{-450}{72} + frac{1}{2}. sin(0)]
[y(0) = 2 + frac{-25}{4} + 0 = 2 - 6.25 = -4.25]
However, the given measurement considers y(t) positive upwards. Therefore, the actual displacement at t = 0 is H(0) = |y(0)| = |-4.25| = 4.25. Thus, the displacement at t = 0 is H(0) = 2 meters.
The initial displacement of the weight from equilibrium at t = 0 is 2 meters upwards. This measurement considers the position of the weight at t = 0 relative to the equilibrium position. The calculation incorporates the given initial conditions of displacement, velocity, and the external force acting on the system. Utilizing the equation of motion for simple harmonic motion allows for the determination of the displacement of the weight at the specific time t = 0.