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Spring Frequency

  • Thread starter jdeakons
  • Start date
3
0
1. Homework Statement
A mass of .88 kg when fastened to the lower end of a light-weight spring and set vibrating up and down is found to have a frequency of 1.8 Hz. Calculate the new frequency of vibration when the mass is decreased to .4 kg.

2. Homework Equations
P = 2 [tex]\pi[/tex] [tex]\sqrt{m/k}[/tex]

3. The Attempt at a Solution
Using K = (m 4 [tex]\pi[/tex]^2 ) / P^2, my answer for K comes out to 112.56.

But, using [tex]\sqrt{k}[/tex] = ([tex]\sqrt{m}[/tex] 2 [tex]\pi[/tex] )/ P, I get 3.2572. But how can that be? The equations are the same, aren't they? Where am I making the mistake?

Thanks.
 

Answers and Replies

458
0
[tex]2\pi f=\sqrt{\frac{k}{m}}[/tex]
 

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