Question

Assume that we have a 3 kg mass attached to a spring with k=5 N/m. (a) What dampener (with what coefficient) would you need to attach to this spring for the system to be critically damped? (b) Assume that you use a dampener with coefficient 2Ns/m, and you will start the system by pulling the mass up by 0.8 m and pushing it downward at a speed of 0.5 m/s. Set up the initial value problem for this situation.

(c) Find the solution to this initial value problem. The coefficients are not pretty, but be patient with it.
(d) Write this solution in the form Re−rhotcos(ωt−δ) (you can leave δ as an inverse trigonometric function). Use this expression and the idea of envelope curves to determine a value of T after which the spring will always stay within 0.05 m of the equilibrium position. A calculator will be required for this part.

Answer 1

In a mass-spring system with a spring constant of 5 N/m and a mass of 3 kg, the dampener needed for critical damping would have a coefficient of approximately 2 Ns/m. The initial value problem for this system involves solving the differential equation 3x'' + 2Ns/mx' + 5 N/mx = 0. The solution to this equation is x(t) = e^(-0.34 Ns/m t)(C1 cos(1.842 Ns/m t) + C2 sin(1.842 Ns/m t)). This solution can be written in the form Re^(-rhot)cos(ωt-δ), where R is the amplitude, ω is the angular frequency, and δ is the phase angle. The envelope curves can be used to determine a value of T after which the spring will always stay within 0.05 m of the equilibrium position.

Step-by-step explanation:

(a) To achieve critical damping in a mass-spring system with a spring constant k and mass m, the damping coefficient b can be calculated using the formula:

b = 2√(mk)

Given that k = 5 N/m and m = 3 kg, we can substitute these values into the formula:

b = 2√(5 N/m × 3 kg)

(b) The initial value problem for this situation can be set up using Newton's second law and the damping force equation:

m&ddot;x + b˙x + kx = 0

Substituting the given values into the equation:

3&ddot;x + 2Ns/m˙x + 5 N/m x = 0

(c) To solve this differential equation, we assume a solution of the form x = Ae^(rt), where A is the amplitude and r is a constant. Substituting this into the equation, we can solve for r: 3r^2 + 2Ns/mr + 5 N/m = 0. Solving this quadratic equation, we find two values for r: r1 ≈ -0.34 Ns/m and r2 ≈ -1.84 Ns/m.

With these values, we can write the general solution to the differential equation as:

x(t) = e^(-0.34 Ns/m t)(C1 cos(1.842 Ns/m t) + C2 sin(1.842 Ns/m t))

(d) To write this solution in the form Re^(-rhot)cos(ωt-δ), we can use the formula:

r = √(b^2 - 4mk)/(2m)

ω = √(4mk - b^2)/(2m)

R = A e^(-bt/2m)

δ = arctan(-2mr/b)

Substituting the given values into these formulas, we can calculate the values of R and δ:

r ≈ -0.34 Ns/m

ω ≈ 1.842 Ns/m

R ≈ Ae^(0.34 Ns/m t)

δ ≈ arctan(-0.364)

Using the expression Re^(-rhot)cos(ωt-δ), we can determine the envelope curves by finding the maximum and minimum values of R. The envelope curves represent the maximum and minimum amplitudes of the oscillation. To find a value of T after which the spring will always stay within 0.05 m of the equilibrium position, we can substitute the value of R and solve for t when the amplitude of R is 0.05 m. This requires the use of a calculator.

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