A 3 kg block is attached to an ideal spring with a force constant k=200 N/m. The block is given an initial velocity in the positive direction of magnitude u= 12 m/s and no initial displacement (xi=0). Find a) the amplitude, b) the phase angle, and c) write an equation for the position as a function of time.
To find the amplitude, phase angle, and equation for the position as a function of time for the block attached to the spring, we can use the principles of simple harmonic motion.
First, let's find the angular frequency (ω) of the system. The angular frequency is given by the formula ω = √(k/m), where k is the force constant and m is the mass of the block.
Given:
k = 200 N/m
m = 3 kg
Substituting these values into the equation, we can calculate the angular frequency:
ω = √(200 N/m / 3 kg) = √(66.67 rad/s^2) ≈ 8.164 rad/s
a) The amplitude (A) of the motion is the maximum displacement from the equilibrium position. In this case, since the block is attached to the spring and there is no initial displacement (xi = 0), the amplitude is simply the initial displacement.
Therefore, the amplitude (A) is 0 meters.
b) The phase angle (ϕ) represents the initial phase of the motion. Since the block is given an initial velocity in the positive direction, the phase angle is 0.
Therefore, the phase angle (ϕ) is 0 radians.
c) The equation for the position as a function of time is given by x(t) = A * cos(ωt + ϕ), where x(t) is the position at time t, A is the amplitude, ω is the angular frequency, and ϕ is the phase angle.
In this case, the equation simplifies to x(t) = 0 * cos(8.164t + 0), which further simplifies to x(t) = 0, since the amplitude (A) is 0.
Therefore, the equation for the position as a function of time is x(t) = 0.
Note: Since there is no initial displacement, the position remains at the equilibrium position throughout the motion. Therefore, x(t) = 0.
a) 2m
b) 43 degrees
c) s(t)=144t^2+26t+9