Undamped, forced oscillation?!?!?
I have this question and can't seem to get the right answer.
Damping is negligible for a 0.130 kg object hanging from a light 6.30 N/m spring. A sinusoidal force with an amplitude of 1.70 N drives the system. At what frequency will the force make the object vibrate with an amplitude of 0.440 m?
I tried the equation A= +/- (f/m) / (W^2-Wo^2) and can't seem to figure out a certain step. Any help would be much appreciated! :)
To solve this problem, we can use the equation for the frequency of a forced oscillation in a system with negligible damping:
f = 1 / (2π) * √(k/m)
Where:
f is the frequency of the forced oscillation
k is the spring constant (6.30 N/m)
m is the mass of the object (0.130 kg)
First, let's calculate the natural frequency of the spring:
ω₀ = √(k/m)
ω₀ = √(6.30 N/m / 0.130 kg)
ω₀ = √(48.4615 N/kg)
ω₀ ≈ 6.958 rad/s
To find the frequency of the forced oscillation, we need to convert the amplitude of the driving force to the angular frequency. Remember that the angular frequency (ω) is related to the frequency (f) by the equation:
ω = 2πf
Given that the amplitude of the driving force is 1.70 N, we can use the equation:
A = F / m * (1 / (√((ω₀^2 - ω^2)^2 + (bω)^2)))
Where:
A is the amplitude of the forced oscillation
F is the amplitude of the driving force
m is the mass of the object
ω₀ is the natural frequency of the spring
b is the damping coefficient (negligible for this problem)
We can rearrange this equation to solve for ω:
(ω₀^2 - ω^2)^2 + (bω)^2 = (F / (m * A))^2
Since bω is negligible in this problem, we can ignore the second term:
(ω₀^2 - ω^2)^2 ≈ (F / (m * A))^2
Expanding the left-hand side:
(ω₀^4 - 2ω₀^2 * ω^2 + ω^4) ≈ (F^2 / (m^2 * A^2))
Now, we can substitute the given values:
(6.958^4 - 2 * 6.958^2 * ω^2 + ω^4) ≈ (1.7^2 / (0.130^2 * 0.440^2))
Now, we can solve this equation numerically to find the value of ω. You can use a graphing calculator or a numerical solver to find the value.