Consider a mass-spring-dashpot system with mass 5kg, a spring which is stretched 3 meters by a force of 10N, and a dashpot which provides 4N resistance for each m/s of velocity. The mass is also acted on by a periodic force 5 cos(ωt) for some number ω.
• For what value of ω (if any) does practical resonance occur?
• If the mass starts at rest, at equilibrium, find a formula for x(t) (in terms of ω), the distance between the mass and the equilibrium point t seconds after motion starts.
To determine whether practical resonance occurs in a mass-spring-dashpot system with a periodic force, we need to find the value of ω that leads to resonance. Practical resonance occurs when the frequency of the periodic force matches the natural frequency of the system.
The natural frequency, ω_0, of a mass-spring-dashpot system can be calculated using the formula:
ω_0 = √(k/m)
where k is the spring constant and m is the mass.
In this case, the spring is stretched by a force of 10N over a distance of 3 meters. The spring constant, k, can be calculated using Hooke's Law:
k = F/x
where F is the force and x is the displacement. Plugging in the values:
k = 10N / 3m = 10/3 N/m
The mass, m, is given as 5kg.
Now we can calculate the natural frequency:
ω_0 = √(k/m) = √((10/3) N/m) / 5kg)
= √((10/3) / 5) (1/√kg)
= √(2/3) (1/√kg)
= √(2/3) (1/√s) (since kg is the same as s^2)
Now, let's find the value of ω for resonance. Resonance occurs when the frequency of the periodic force is equal to the natural frequency of the system. The frequency of the periodic force is given by ω = 2π/T, where T is the time period of the force.
Since the periodic force is given as 5 cos(ωt), we see that the force has a frequency of ω. Therefore, for resonance to occur, we need ω = ω_0.
So, the value of ω for practical resonance to occur in this system is ω = √(2/3) (1/√s).
Next, let's find a formula for x(t), the distance between the mass and the equilibrium point at time t, given that the mass starts at rest.
The equation of motion for the mass-spring-dashpot system is:
m * (d^2x/dt^2) + b * (dx/dt) + k * x = F_0 * cos(ωt)
Where m is the mass, b is the damping coefficient (in this case, b = 4N/(m/s) = 4 kg/s), k is the spring constant, x is the displacement from equilibrium, F_0 is the amplitude of the periodic force, and ω is the angular frequency.
Since the mass starts at rest (dx/dt = 0 at t = 0), we can set dx/dt = 0 in the equation of motion. This simplifies the equation to:
m * (d^2x/dt^2) + k * x = F_0 * cos(ωt)
To find the particular solution for x(t), we assume x(t) has the form:
x(t) = A * cos(ωt)
where A is the amplitude of the displacement.
Differentiating twice with respect to time, we find:
dx/dt = -A * ω * sin(ωt)
d^2x/dt^2 = -A * ω^2 * cos(ωt)
Substituting these into the equation of motion:
m * (-A * ω^2 * cos(ωt)) + k * A * cos(ωt) = F_0 * cos(ωt)
Simplifying and factoring out cos(ωt), we get:
(A * k / m - A * ω^2) * cos(ωt) = F_0 * cos(ωt)
Since cos(ωt) is non-zero, we can divide both sides of the equation by cos(ωt):
A * k / m - A * ω^2 = F_0
Rearranging the equation:
A * (k / m - ω^2) = F_0
Solving for A:
A = F_0 / (k / m - ω^2)
Therefore, the formula for x(t) is:
x(t) = (F_0 / (k / m - ω^2)) * cos(ωt)
Note that this expression assumes that the angular frequency, ω, is not equal to the natural frequency, ω_0. When ω = ω_0, the system exhibits resonance and has a different solution.