Find Spring Constant k Using Mass & Time Values

AI Thread Summary
To find the spring constant k using mass and time values, the relevant formula is T = 2π√(m/k), which can be rearranged to T² = (4π²m)/k. The discussion clarifies that the period T is the time for one complete oscillation, and if the average time for 10 vibrations is 8.86 seconds, the period can be calculated by dividing this time by 10. This yields a period of 0.886 seconds for one oscillation. Understanding the relationship between mass, time, and the spring constant is essential for solving the problem effectively.
HELP_786
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how would i go about finding k with given values for mass and time?

i have seen these equations over and over...but i feel like I am missing something very important that i need..please help me

T=2π√(m/k))

F=-kx
 
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T= 2 \pi \sqrt{\frac{m}{k}}

\Rightarrow T^2=\frac{4 \pi^2 m}{k}

Can you rearrange for k now?
 
Last edited:
o I am sorry i meant how would I go about finding T(period) with given values for time and mass...i forgot to proof read :redface:

those are the formulas i keep seeing everywhere

for example if i had .10kg and 8.86 seconds for the time to find the period would i do this:

T2=(4π2m)/((4π2m)/(8.86))

and then at the end take the square root

?
 
How exactly was the time measured?
 
in seconds

8.86 seconds would be the average time for 10 vibrations in a simple harmonic motion
 
HELP_786 said:
8.86 seconds would be the average time for 10 vibrations in a simple harmonic motion


Period(T) is the time taken for one oscillation/vibration.


10 oscillations/vibrations take 8.86 seconds. How much time would 1 oscillation take?
 
OH! so just divide 8.86s by 10 rite??
 
HELP_786 said:
OH! so just divide 8.86s by 10 rite??

That should yield the time period, yes.
 
i don't believe i didnt realize that! thanku so much! ...im thinking too much lol
 
  • #10
HELP_786 said:
i don't believe i didnt realize that! thanku so much! ...im thinking too much lol

You just need to think about what it is you want to find.
 
  • #11
thanx again! =D
 
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