How Do You Find the Total Energy in Simple Harmonic Motion?

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Homework Help Overview

The discussion revolves around calculating the total energy of a mass-spring system undergoing simple harmonic motion. The original poster presents a problem involving a 45.0-g object attached to a spring with a specified force constant and amplitude, seeking to determine the total energy in millijoules.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the force constant and spring constant, questioning whether they are equivalent. There are discussions about unit conversions, particularly regarding amplitude from centimeters to meters, and how this affects the energy calculation. Some participants attempt to clarify the units associated with energy and the formula used.

Discussion Status

The conversation is ongoing, with participants providing guidance on unit conversions and the implications of using different units in the energy formula. There is recognition of the need to convert amplitude to meters to obtain the correct units for energy, but no consensus has been reached on the numerical value of the energy calculation.

Contextual Notes

Participants are navigating through the constraints of the problem, including the requirement to express the final answer in millijoules and the implications of using different units for amplitude in the energy formula.

Erenjaeger
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Homework Statement


A 45.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 6.00 cm on a frictionless, horizontal surface.
a) find the total energy of the system (mJ)[/B]

Homework Equations


1/2KA^2
[/B]

The Attempt at a Solution


Is the force constant the same as the spring constant 'K' here ? or will i need to determine 'K' by using Hooke's Law? f=-kx and then rearrange for -k=f/x?
also does the A have to be converted into m or can i just plug in 6.00cm?[/B]
 
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Erenjaeger said:
Is the force constant the same as the spring constant 'K' here ?
Yes.
or will i need to determine 'K' by using Hooke's Law? f=-kx and then rearrange for -k=f/x?
There is no way to determine K this way by using information given in the problem.
also does the A have to be converted into m or can i just plug in 6.00cm?
Try plugging the numbers into the formula along with the units and see what units you end up with for the energy.
 
TSny said:
Yes.
There is no way to determine K this way by using information given in the problem.
Try plugging the numbers into the formula along with the units and see what units you end up with for the energy.
n⋅m cm2 ??
 
Erenjaeger said:
n⋅m cm2 ??
Not quite. Note that K has units of Newtons per meter.
 
TSny said:
Not quite. Note that K has units of Newtons per meter.
yeah so n/m but then then amplitude has units of cm with the ^2 there so unless i convert amplitude into m maybe??
 
If you can get the energy in Joules, then it won't be hard to convert to mJ. Can you relate the Joule to the Newton and the meter?
 
TSny said:
If you can get the energy in Joules, then it won't be hard to convert to mJ. Can you relate the Joule to the Newton and the meter?
right so the answer i was getting was in joules because joules is n/m or kg⋅m^2⋅s^2 so is mJ millijoules or megajoules ?
 
Erenjaeger said:
right so the answer i was getting was in joules because joules is n/m or kg⋅m^2⋅s^2
J ≠ N/m. Did you mean to write N⋅m rather than N/m?
Also, did you mean to write kg⋅m^2/s^2 instead of kg⋅m^2⋅s^2?
so is mJ millijoules or megajoules ?
mJ stands for milliJoule. MegaJoule would be MJ.
 
TSny said:
J ≠ N/m. Did you mean to write N⋅m rather than N/m?
Also, did you mean to write kg⋅m^2/s^2 instead of kg⋅m^2⋅s^2?
mJ stands for milliJoule. MegaJoule would be MJ.
yeah i thought K had a unit of Newtons per meter ?
right so they want the answer i got but in milliJoule?
 
  • #10
Erenjaeger said:
yeah i thought K had a unit of Newtons per meter ?
Yes, K has units of N/m. But what about energy? How would you express the Joule in terms of N and m?
right so they want the answer i got but in milliJoule?
Yes, they want the answer in milliJoules.
 
  • #11
In the formula E = (1/2)KA2 suppose you used N/m for the units of K and cm for the unit of A. What would you get for the units for E? Would you end up with units corresponding to Joules?
 
  • #12
TSny said:
In the formula E = (1/2)KA2 suppose you used N/m for the units of K and cm for the unit of A. What would you get for the units for E? Would you end up with units corresponding to Joules?
would you get N⋅m⋅cm2 ??
 
  • #13
Erenjaeger said:
would you get N⋅m⋅cm2 ??
No. The units for K are N/m, not N⋅m.
 
  • #14
TSny said:
No. The units for K are N/m, not N⋅m.
dude i am so lost, can you not just explain it to me so i can learn about what youre talking about?
 
  • #15
If you use K = 40.0 N/m and A = 6.00 cm in the formula E = (1/2)KA2, what do you get for an answer for E (including units)?
 
  • #16
TSny said:
If you use K = 40.0 N/m and A = 6.00 cm in the formula E = (1/2)KA2, what do you get for an answer for E (including units)?
120 N/m⋅cm2 since the amplitude is given in cm.??
 
  • #17
Erenjaeger said:
120 N/m⋅cm2 since the amplitude is given in cm.??
Yes, that's right. Does N/m⋅cm2 represent Joules?

[EDIT: Sorry, I was only looking at the units. You got the units right, but the numerical value is not 120.]
 
  • #18
TSny said:
Yes, that's right. Does N/m⋅cm2 represent Joules?
doesnt a joule equal a N⋅m ??
 
  • #19
Erenjaeger said:
doesnt a joule equal a N⋅m ??
Yes, good. So using 6 cm for A does not yield N⋅m (or Joules) for E.
 
  • #20
TSny said:
Yes, good. So using 6 cm for A does not yield N⋅m (or Joules) for E.
oh yeah so for N⋅m would i just convert amplitude to meters? but then I am still left with N/m⋅cm2 does the cm2 part affect anything?
 
  • #21
Erenjaeger said:
oh yeah so for N⋅m would i just convert amplitude to meters? but then I am still left with N/m⋅cm2 does the cm2 part affect anything?
If you convert the amplitude to meters, then you will not have any cm in your answer.
 
  • #22
TSny said:
If you convert the amplitude to meters, then you will not have any cm in your answer.
oh true, so by converting the amplitude to m, I am left with N.m which is the units for joules right?
 
  • #23
Erenjaeger said:
oh true, so by converting the amplitude to m, I am left with N.m which is the units for joules right?
Yes, that's right.
 
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