Simple Harmonic Motion-oscillation on an ideal spring

  • Thread starter MissEuropa
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  • #1
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Homework Statement



When a 0.870kg mass oscillates on an ideal spring, the frequency is 1.33Hz.
a) What will the frequency be if 0.290kg are added to the original mass? Try to solve this problem without finding the force constant of the spring.

b) What will the frequency be if 0.290kg are subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Homework Equations


f=(1/T) or1/2π*√k/m

The Attempt at a Solution


I'm at a loss as to how to approach this without using the spring constant k.
My attempt at a solution so far anyway:
frequency= c/m where c is a constant and m is the mass.
Solving for c in the equation 1.33 Hz=c/0.870kg I found c= 1.151
Then I divided c (1.151) by the original mass plus the added mass.
1.151/(0.870kg + 0.290kg) = .992

But this is incorrect, I want to attempt to do the prompt as instructed without using k but I'm at a loss currently. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
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I think your method was indirectly solving for k hidden inside of that constant you chose.

You can multiply the new mass into the equation
[tex] \omega_1=\sqrt{\frac{k}{m_1}}[/tex]
to get
[tex]\omega_1 \sqrt{m_2}=\sqrt{k\frac{m_2}{m_1}}[/tex]
and you can subsitute in the solution for m2 from the equation
[tex]\omega_2=\sqrt{\frac{k}{m_2}}[/tex]
solved for m2
[tex]\sqrt{m_2}=\frac{\sqrt{k}}{\omega_2}[/tex]

That should lead you on your way, the k's will fall out of the equation.

Oh and just in case the notation I used is new..
[tex]\omega=2\pi f[/tex]
 
Last edited:
  • #3
349
1
You are on the right track but f = c/√M
Use this to find c then substitue the new masses
 
  • #4
20
0
I think your method was indirectly solving for k hidden inside of that constant you chose.

You can multiply the new mass into the equation
[tex] \omega_1=\sqrt{\frac{k}{m_1}}[/tex]
to get
[tex]\omega_1 \sqrt{m_2}=\sqrt{k\frac{m_2}{m_1}}[/tex]
and you can subsitute in the solution for m2 from the equation
[tex]\omega_2=\sqrt{\frac{k}{m_2}}[/tex]
solved for m2
[tex]\sqrt{m_2}=\frac{\sqrt{k}}{\omega_2}[/tex]

That should lead you on your way, the k's will fall out of the equation.

Oh and just in case the notation I used is new..
[tex]\omega=2\pi f[/tex]
Awesome, that did help a lot. I think I was not going down that route because I didn't consider that the k's would cancel out.
:-) Thanks again!
 

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