A particle of mass m = 0.35 kg is oscillating on the end of a spring with amplitude 0.32 m. It is at its equilibrium position (x=0), moving in the positive x-direction, at t=0, and 0.18 s later it is at x=0.25 m.
a. What is the lowest possible value for the spring constant?
b. What is the second lowest possible value for the spring constant?
To find the lowest possible value for the spring constant, we can use the equation for the period of a mass-spring system:
T = 2π√(m/k),
where T is the period, m is the mass of the particle, and k is the spring constant.
Since the particle completes one oscillation in 0.18 seconds, we can determine the period T by dividing 0.18 by the number of oscillations (1). This gives us T = 0.18 s.
Rearranging the equation, we have:
k = (4π²m) / T².
Substituting the known values:
k = (4π² * 0.35 kg) / (0.18 s)².
Now we can calculate the value of k:
k = (4 * 3.14² * 0.35) / (0.18)².
k ≈ 94.61 N/m.
Therefore, the lowest possible value for the spring constant is approximately 94.61 N/m.
To find the second lowest possible value for the spring constant, we need to consider that the amplitude affects the motion of the particle. The equation for the period in terms of amplitude is:
T = 2π√(m/k) * √(1 - (A² / x₀²)),
where A is the amplitude of the motion and x₀ is the initial displacement from equilibrium.
We know that the amplitude A is 0.32 m and the displacement x₀ is 0.25 m.
Rearranging the equation, we have:
k = (4π²m) / (T² * (1 - (A² / x₀²))).
Substituting the known values:
k = (4π² * 0.35 kg) / (0.18 s)² * (1 - ((0.32 m)² / (0.25 m)²)).
Calculating the value of k:
k = (4 * 3.14² * 0.35) / (0.18)² * (1 - (0.32² / 0.25²)).
k ≈ 40.54 N/m.
Therefore, the second lowest possible value for the spring constant is approximately 40.54 N/m.