A 0.450 kg mass attached to a horizontal spring is displaced 11.4 cm from rest and then released. If the mass oscillates 7.00 times in 11.5 s, what is the spring constant in the spring? 0.167 N/m 6.58 N/m 0.134 N/m 14.6 N/m
Question
A 0.450 kg mass attached to a horizontal spring is displaced 11.4 cm from rest and then released. If the mass oscillates 7.00 times in 11.5 s, what is the spring constant in the spring? 0.167 N/m 6.58 N/m 0.134 N/m 14.6 N/m
Solution
To solve this problem, we need to use the formula for the period of a simple harmonic oscillator, which is T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.
First, we need to find the period of the oscillation. We know that the mass oscillates 7.00 times in 11.5 s, so the period T is 11.5 s / 7.00 = 1.643 s.
Next, we can rearrange the formula to solve for k: k = m / (T/2π)^2.
Substituting the given values: k = 0.450 kg / (1.643 s / 2π)^2 = 6.58 N/m.
So, the spring constant is 6.58 N/m.
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