If a 20 kg of mass, m, is attached to the above spring in problem 20 (spring constant k), what will the frequency of oscillation, f, be?
A spring is 0.10 m long when no force is applied to it. When 4 N pull is applied to the spring, it stretches to a total length of 0.15 m. What is the spring constant for this spring?
- electron1Lv 76 months agoFavorite Answer
f = 1/(2 * π) * √(k/m)
The first step is solve problem 20 to determine the spring constant.The unit of the spring constant is N/m The first step is to determine the increase of the length.
∆ L = 0.15 – 0.10 = 0.05 meter
k = 4 ÷ 0.05 = 80 N/m
Now we can use the spring constant to calculate the frequency.
k/m = 80/20 = 4
√(k/m) = 2
f = 1/2π * 2
The frequency is 1/π. This is approximately 0.32 Hz. I hope this is helpful for you.
- NCSLv 76 months ago
spring constant k = force / elongation = 4N / 0.05m = 80 N
plug in values and get
f = 0.32 Hz
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