When released from rest, the block drops 0.127 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.
Help would be much appreciated!!
Help would be much appreciated!!
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0.127m describes twice the amplitude of the motion -- it will return to 0m (and 0m/s) at the top of the movement as well. Therefore, at the new equilbrium, the force in the spring will equal the weight of the object, or
kx = mg
k * 0.127m/2 = 0.463kg * 9.8m/s²
k = 71 N/m
b) ω = √(k/m) = √(71N/m / 0.463kg) = 12.4 rad/s
kx = mg
k * 0.127m/2 = 0.463kg * 9.8m/s²
k = 71 N/m
b) ω = √(k/m) = √(71N/m / 0.463kg) = 12.4 rad/s