How Long Does It Take for a Mass to Oscillate Between Two Points on a Spring?

AI Thread Summary
A vertical spring stretches 9.6 cm with a 1.3 kg block, which is then displaced an additional 5.0 cm and released. The problem involves calculating the time for the mass to oscillate between two points, 0.025 m above and below the equilibrium position. The correct angular frequency is determined to be 10.1, leading to the equations X1 and X2 for the oscillation. The initial calculations led to confusion regarding the use of radians versus degrees, which affected the time values. Ultimately, the correct times are t1 = 0.1 seconds and t2 = 0.2 seconds, confirming a total time change of 0.1 seconds.
shasanni
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Homework Statement



A vertical spring stretches 9.6 cm when a 1.3kg block is hung from its end.This block is then displaced an additional 5.0 cm downward and released from rest. Calculate the time it takes the object to go from a point 0.025 m below its equilibrium position to a point 0.025 m above it.


Homework Equations


X1= Acos(Wt1)
X2=Acos(Wt2)
A represents the amplitude, which is 0.05 m. The W represents the angular frequency, which is 10.1




The Attempt at a Solution


This is how I sent things up:
0.025=0.05cos(10.1t1)
-0.025=0.05cos(10.1t2)
and then I would have subtracted the two times, but I am not getting the correct values for t1 and t2. What am I doing wrong??
Btw, the answer is supposed to be 0.1s. T2 is 0.2s and T1 is 0.1s. ?
 
Last edited:
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What did you get for t1 and t2?

ehild
 
This are the values I get:
t1= 0.507
t2=-.0507
 
cos(ωt1)=0.5 corresponds to ωt1=π/3.
cos(ωt2)=-0.5 corresponds to ωt2=π-π/3.
What are t1 and t2 then?

ehild
 
i think I know what the issue is. I have been using radians, not degrees!
t1 should 0.1 seconds and t2 should be 0.2 s. The change in time should be 0.1 seconds, which is the answer.
 
Well done! take care of radians and degrees in the future.

ehild
 
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