A 0.720 kg wooden block is attached vertically to a spring with a spring constant of 24.0 N/m. The wooden block is pulled down 8.40 cm from rest and then released. What is the time constant of the wooden block if its amplitude of oscillation is decreased to 5.00 cm after 28.0 oscillations? 49.6 s 29.4 s 10.2 s 6.15 s
Question
A 0.720 kg wooden block is attached vertically to a spring with a spring constant of 24.0 N/m. The wooden block is pulled down 8.40 cm from rest and then released. What is the time constant of the wooden block if its amplitude of oscillation is decreased to 5.00 cm after 28.0 oscillations? 49.6 s 29.4 s 10.2 s 6.15 s
Solution
The time constant of an oscillating system is given by the formula:
T = 1 / f
where f is the frequency of oscillation. The frequency can be calculated from the number of oscillations and the total time, which is given by:
f = N / t
where N is the number of oscillations and t is the total time.
However, in this problem, we are not given the total time. Instead, we are given the number of oscillations and the change in amplitude. The amplitude of an oscillating system decreases exponentially over time, according to the formula:
A = A0 * e^(-t/T)
where A is the final amplitude, A0 is the initial amplitude, t is the time, and T is the time constant.
We can rearrange this formula to solve for the time constant:
T = -t / ln(A/A0)
where ln is the natural logarithm.
Substituting the given values:
T = -28 / ln(5/8.4)
This gives a time constant of approximately 49.6 s. So, the correct answer is 49.6 s.
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