A spring with spring constant 14.0 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to rest. It is then pulled down 8.00 cm and released.
What is the time constant if the ball's amplitude has decreased to 2.90 cm after 31.0 oscillations?
"Time constant" usually refers to nonoscillatory motion, with an exponential decay. Do you want the "damping constant" that relates the friction force to velocity? Or the constant term that describes the exponential decay of the amplitude of the oscillations?
The solution to the equation of motion is of the form
X = 8 exp(-t/T) cos wt
w = sqrt(k/m) = 5.34 rad/s
(Actually, the damping reduces the frequency of oscillation slightly)
Period = 2 pi/w = 1.18 s
31 oscillations require 36 seconds.
exp(-36/T) = 2.9/8.0 = 0.363
T = 35 seconds is an effective time constant for decay.
To find the time constant, we can use the formula:
T = (t × N) / n
Where:
T is the time constant,
t is the time elapsed,
N is the number of oscillations, and
n is the natural logarithm of the amplitude ratio.
Given:
t = 31.0 oscillations
N = 31.0 oscillations
n = ln(A1/A2)
A1 = Initial amplitude = 8.00 cm
A2 = Final amplitude = 2.90 cm
First, let's calculate the value of n:
n = ln(A1 / A2)
= ln(8.00 cm / 2.90 cm)
≈ ln(2.7586)
≈ 1.0138
Now, substitute the given values into the formula for the time constant T:
T = (t × N) / n
= (31.0 oscillations × 31.0 oscillations) / 1.0138
≈ 959.37 oscillations^2
Therefore, the time constant is approximately 959.37 oscillations^2.
To find the time constant, we need to first find the damping constant. The damping constant can be found using the formula:
ζ = (2πn) / (ln(a1 / a2))
Where:
ζ is the damping constant
n is the number of full periods elapsed
a1 is the initial amplitude
a2 is the final amplitude
In this case, n = 31, a1 = 8.00 cm, and a2 = 2.90 cm.
Let's substitute these values into the formula:
ζ = (2π * 31) / (ln(8.00 / 2.90))
Now, let's calculate ζ:
ζ ≈ (2π * 31) / (ln(2.76))
Next, we can calculate the time constant using the formula:
τ = 1 / (ζ * ω)
Where:
τ is the time constant
ζ is the damping constant
ω is the angular frequency
The angular frequency can be determined using the formula:
ω = √(k / m)
Where:
k is the spring constant
m is the mass attached to the spring
In this case, k = 14.0 N/m and m = 490 g (which is equal to 0.49 kg).
Let's calculate ω:
ω = √(14.0 / 0.49)
Now, let's calculate τ:
τ ≈ 1 / (ζ * ω)
After finding ζ and ω, we can substitute them into the formula:
τ ≈ 1 / (ζ * ω)
Finally, evaluate the expression to find the time constant.