- #1
Torquenstein101
- 11
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An object of unknown mass is hung on the end of an unstretched spring and is released from rest. If the object falls 3.42 cm before first coming to rest, find the period of the motion.
Doing a summation of forces, i see that the force constant k times the displacement y is equal to the weight of the object. From this I see that (k/m)=(g/y). That means the angular frequency (omega) is equal to the square root of (g/y). Knowing that the period of the oscillation T is the inverse of the frequency, i get T=(2pie)/omega which is then (2pie) divided the square root of (g/y). Solving for T, i get 0.371 seconds. However, this doesn't seem to be the right answer. The correct answer is 0.262 seconds, but I just don't see how that answer was obtained. Any help is greatly appreciated. Thanks.
Doing a summation of forces, i see that the force constant k times the displacement y is equal to the weight of the object. From this I see that (k/m)=(g/y). That means the angular frequency (omega) is equal to the square root of (g/y). Knowing that the period of the oscillation T is the inverse of the frequency, i get T=(2pie)/omega which is then (2pie) divided the square root of (g/y). Solving for T, i get 0.371 seconds. However, this doesn't seem to be the right answer. The correct answer is 0.262 seconds, but I just don't see how that answer was obtained. Any help is greatly appreciated. Thanks.