for a simple harmonic experiment using a spring-mass system -near the bottom of
ID: 1974151 • Letter: F
Question
for a simple harmonic experiment using a spring-mass system-near the bottom of the spring, the spring only supports the mass of the mass hanger which is (50g). Assuming its k value is 12N/m, how much does it want to stretch(distance)?
-Near the top, the spring has the mass hanger(50g)+ its own weight(75g), now how much does it want to stretch/
- The springs we had had a changing k; it was higher at the top than at the bottom( it looked tapered), why do you think it was that way?
-what do you think would have happened if you put it upside down?( what would the spring look like, not how it would change the experiment)
Explanation / Answer
--- kx = 50*10-3 *9.8
x = .49/12 = 0.0408 m
x = 4.08 cm ... ANS
---- kx = (50+75)10-3 *9.8
x = 1.225/12 = .102 m
x = 10.2 cm
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