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Simple Harmonic Motion confusion need help!

  • Thread starter knottlena
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1. A 14.7g mass is attached to a frictionless horizontal spring. After the spring was stretched a certain distance and released, its maximum speed was found to be 4.22 m/s. If the speed at x = -2.84 cm is 1.63 m/s, what is the frequency of the oscillating spring?

2. PEe = kx^2/2 E=KE + PE E=mV^2/2 + kx^2/2 (E is conserved) f = 1/3.14 sq. rt of k/m


3. (.0147kg)(-.0284m)^2/2 = .000005928 J

I don't know how to get (k) and honestly have no idea where to start this problem.
 

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  • #2
rock.freak667
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3. (.0147kg)(-.0284m)^2/2 = .000005928 J

I don't know how to get (k) and honestly have no idea where to start this problem.


If you know that

E = KE + PE = ½mv2 + ½kx2

When v is maximum what is x ?

That will give you the total energy in the system.

Then you know v for a given x, so just put that into the equation again to get k.
 
  • #3
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If you know that

E = KE + PE = ½mv2 + ½kx2

When v is maximum what is x ?

That will give you the total energy in the system.

Then you know v for a given x, so just put that into the equation again to get k.

This is where I am just going around in circles. I have Vmax as 4.22 m/s. That is what the question gives, but without (k) I can't find (x) for the Vmax. I can't find PEmax without the (x). The only x value that I have is the -2.84 cm and that is when the spring is now being compressed not stretched. What am I missing here? Why is this so hard?

If I do KE= 1/2(.0147kg)(1.63m/s)^2 = .0195 J or is it N? PE= 1/2 (k)(.-0284m)= PE= (-.0142)(k) Now is PE supposed to be interconvertable with KE?
 
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  • #4
rock.freak667
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This is where I am just going around in circles. I have Vmax as 4.22 m/s. That is what the question gives, but without (k) I can't find (x) for the Vmax. I can't find PEmax without the (x). The only x value that I have is the -2.84 cm and that is when the spring is now being compressed not stretched. What am I missing here? Why is this so hard?

If I do KE= 1/2(.0147kg)(1.63m/s)^2 = .0195 J or is it N? PE= 1/2 (k)(.-0284m)= PE= (-.0142)(k) Now is PE supposed to be interconvertable with KE?
with Etotal = KE + PE, as KE increases PE decrease, so when KE is maximum PE is zero. So using Vmax you can get Etotal.

Then you use the second condition of v=1.63 m/s when x= -2.84 cm to get k.
 
  • #5
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with Etotal = KE + PE, as KE increases PE decrease, so when KE is maximum PE is zero. So using Vmax you can get Etotal.

Then you use the second condition of v=1.63 m/s when x= -2.84 cm to get k.
Thank you for getting back to me so quickly on these, I appreciate it. I am trying so hard. I have gained a few more greys on my head for all this. Anyway...

Is KE interchangable with PE here? If that is so then I would take KE=mV^2/2 = .0147kg(1.63m/s)^2/2 = .0195 and KE is interchangeable with PE then I would do PE = kx^2/2 = .0195J = (k)(-.0284m)^2/2 = .0195(2) = (k)(.000807m) = .039/.000807m = k = 48.3 N/m. Now am I supposed to do (PEmax = KEmax) = (kA^2/2 = mV^2/2) = A= the sq. rt. of mVmax^2/k to get the A for the vmax? If I do then I would get this: A = the sq. rt of (.0147kg)(4.22m/s)^2/48.3 N/m = .0736m.

I thought that PEmax=E? If that is the case I got .1308J. If I use the equation E= 1/2 mV^2 + 1/2 kx^2, plugging in the numbers from above I get E= .2617 J. Now would I have to do PE= E-KE to get .2422 J? I don't see how E works into the equation of f = 1/2(3.14) sq. rt of k/m or which E is the right E...I can't use f = 1/T because I don't have time...wait a minute...could I do T = 2(3.14) sq. rt of m/k and then do 1/T to get the frequency? Do I need E in order to find a new (k)? I thought that (k) always stayed the same?
Say I use E = .1309J then I would get k = 2(.1309)/.0736 = 3.56 (seems like a small k #). If I use E = .2617 I get k = 2(.2617)/.0736 = 7.114. I am confused as to how my (k) can go from 48.3 to either of these #'s. Am I any closer to the right answer yet?
 
  • #6
rock.freak667
Homework Helper
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Is KE interchangable with PE here? If that is so then I would take KE=mV^2/2 = .0147kg(1.63m/s)^2/2 = .0195 and KE is interchangeable with PE then I would do PE = kx^2/2 = .0195J = (k)(-.0284m)^2/2 = .0195(2) = (k)(.000807m) = .039/.000807m = k = 48.3 N/m. Now am I supposed to do (PEmax = KEmax) = (kA^2/2 = mV^2/2) = A= the sq. rt. of mVmax^2/k to get the A for the vmax? If I do then I would get this: A = the sq. rt of (.0147kg)(4.22m/s)^2/48.3 N/m = .0736m.
During the entire motion, KE is converted to PE and vice versa for the the other part of the motion.

So when KE is max, PE is zero and when PE is max KE is zero.

From E = KE + PE, when v = 4.22 (max KE) the equation becomes

E = KEmax, so you can sub in the values and get the value for E.

During any moment that is not max KE or PE, the equation is

E = KE + PE = 0.5mv2 + 0.5kx2

you have E from the first part, then they gave you v for some x. (so you can find k)

Now when you get k, you know that ω=√(k/m) where ω = 2π/T.
 

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