Amplitude vs Location when KE=PE?

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a mass-spring system undergoing simple harmonic motion. The original poster is tasked with finding the amplitude of vibration and the position where kinetic energy equals potential energy. There is confusion regarding the differences between the two parts of the question.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between kinetic and potential energy in the context of oscillation, questioning how the amplitude relates to the position where KE equals PE.
  • Some participants attempt to derive expressions for energy conservation and the conditions under which KE and PE are equal.
  • There is a discussion about the total energy of the system and how it remains constant throughout the motion.
  • Several participants express uncertainty about the calculations leading to the position where KE equals PE, particularly regarding the correct interpretation of energy equations.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance without reaching a consensus. Some participants suggest re-evaluating the steps taken in calculations, while others encourage a shift in perspective regarding energy relationships. There is recognition of the need to clarify fundamental concepts related to energy conservation in harmonic motion.

Contextual Notes

Participants note that the total energy remains constant and question how this affects the calculations for the position where KE equals PE. There is also mention of potential confusion stemming from the algebraic steps taken in deriving the position.

bob tran
Messages
17
Reaction score
0

Homework Statement


A 0.50-kg object is attached to an ideal massless spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
(a) What is the amplitude of vibration?
(b) At what location are the kinetic energy and the potential energy of the system the same?
Correct Answer: a) 0.24 m, b) 0.17 m
I'm not sure how part A is different from part B.

Homework Equations


Part A:
<br /> \dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\<br /> \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2<br />

The Attempt at a Solution


Part A:
<br /> \dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\<br /> A=\sqrt{\dfrac{mv^2}{k}}\\<br /> A=\sqrt{\dfrac{(.5)(1.5)^2}{20}}=0.24 \ \texttt{m}<br />
Part B:
Same thing..?
<br /> \dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2\\<br /> x=\sqrt{\dfrac{mv^2}{k}}\\<br /> x=\sqrt{\dfrac{(.5)(1.5)^2}{20}}=0.24 \ \texttt{m}<br />
 
Physics news on Phys.org
x = 0.24m being the amplitude means that the kinetic energy at that point is clearly zero. For part B, remember that the total energy remains the same.
 
  • Like
Likes   Reactions: bob tran
JeremyG said:
x = 0.24m being the amplitude means that the kinetic energy at that point is clearly zero. For part B, remember that the total energy remains the same.
Hmm.. I still don't understand how this would get 0.17 m as an answer for part B.
 
If the total energy remains the same throughout the motion, at the point where KE=PE, what is the expression of the PE/KE in terms of the total energy E?
 
  • Like
Likes   Reactions: bob tran
JeremyG said:
If the total energy remains the same throughout the motion, at the point where KE=PE, what is the expression of the PE/KE in terms of the total energy E?
\dfrac{PE}{KE}=1 \ ?\\<br /> E = 2PE = 2KE \ ?<br />
 
Last edited:
Yes, and you already know the total energy, correct? Substitute in the expression for PE and you'll be able to solve for the position.
 
  • Like
Likes   Reactions: bob tran
JeremyG said:
Yes, and you already know the total energy, correct? Substitute in the expression for PE and you'll be able to solve for the position.
Isn't it still 0.24 m?
<br /> E = PE + KE\\<br /> E = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\<br /> 2PE = \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\<br /> 2(\dfrac{1}{2}kA^2)= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\<br /> kA^2= \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2\\<br /> kA^2-\dfrac{1}{2}mv^2= \dfrac{1}{2}kx^2\\<br /> \sqrt{\dfrac{2(kA^2-\dfrac{1}{2}mv^2)}{k}}=x\\<br /> \sqrt{\dfrac{2((2)(0.24)^2-\dfrac{1}{2}(.5)(1.5)^2)}{20}}=x\\<br /> 0.24 \ \texttt{m} = x<br />
 
No, look through your steps once again. In particular, what does your 4th step say? That the total energy is 2 times of 1/2kA2? How is that possible? The concept being tested here is once again, that the total energy is conserved throughout the motion.
 
  • Like
Likes   Reactions: bob tran
Another hint for you though. Instead of thinking in terms of E = 2PE = 2KE, why not think of it in terms of KE = PE = 1/2E?

Mathematically equivalent but I hope this makes it clearer with a shift of perspective
 
  • Like
Likes   Reactions: bob tran
  • #10
JeremyG said:
Another hint for you though. Instead of thinking in terms of E = 2PE = 2KE, why not think of it in terms of KE = PE = 1/2E?

Mathematically equivalent but I hope this makes it clearer with a shift of perspective
Still don't see it. :headbang:
The only way I got an answer that rounds to 0.17 m was by doing:
<br /> E=PE+KE\\<br /> E=\dfrac{1}{2}mv^2+\dfrac{1}{2}kA^2=1.125 \ \texttt{J}<br />
And then using that value here:
<br /> \dfrac{E}{2}=PE+\dfrac{KE}{2}\\<br /> E=2PE+KE\\<br /> E=2(\dfrac{1}{2}kx^2)+\dfrac{1}{2}mv^2\\<br /> E=kx^2+\dfrac{1}{2}mv^2\\<br /> E-\dfrac{1}{2}mv^2=kx^2\\<br /> \sqrt{\dfrac{E-\dfrac{1}{2}mv^2}{k}}=x\\<br /> \sqrt{\dfrac{1.125-\dfrac{1}{2}(.5)(1.5)^2}{20}}=x\\<br /> 0.167705 \ \texttt{m}=x\\<br /> 0.17 \ \texttt{m} \approx x<br />
But, I am not sure if that is correct and why we are allowed to do \dfrac{E}{2}=PE+\dfrac{KE}{2}\\, let alone why we should use this step at all to get the correct answer.
 
Last edited:
  • #11
bob tran said:
\dfrac{PE}{KE}=1 \ ?\\<br /> E = 2PE = 2KE \ ?<br />
I think you need to see that TOTAL energy E =( PE at displacement x) + (KE at displacement x)
When displacement x = 0,...PE =0 and KE is max
When displacement x = A,... KE = 0 and PE is max
so it is correct to say that E = PEmax
or E = KEmax
 
  • Like
Likes   Reactions: bob tran
  • #12
bob tran said:
<br /> \dfrac{E}{2}=PE+\dfrac{KE}{2}
Where did that come from?
You know the total energy. How much PE is there when there's equal KE and PE? What extension does that give?
 
  • Like
Likes   Reactions: bob tran
  • #13
lychette said:
I think you need to see that TOTAL energy E =( PE at displacement x) + (KE at displacement x)
When displacement x = 0,...PE =0 and KE is max
When displacement x = A,... KE = 0 and PE is max
so it is correct to say that E = PEmax
or E = KEmax
Still not sure how I use this for the problem. :frown:
haruspex said:
Where did that come from?
You know the total energy. How much PE is there when there's equal KE and PE? What extension does that give?
\dfrac{E}{2}=PE? But, what exactly am I supposed to do with that? When I substitute this into the equation, it still get 0.24 m as the answer. It seems like I don't understand something fundamental about this or I keep substituting in the wrong thing.
 
  • #14
bob tran said:
\dfrac{E}{2}=PE? But, what exactly am I supposed to do with that? When I substitute this into the equation, it still get 0.24 m as the answer. It seems like I don't understand something fundamental about this or I keep substituting in the wrong thing.
I see nowhere in this thread where you have done that correctly.
In your post #7, there is an error going from your second line to the third. This path was not going to work anyway because at the end you still have a v in there, and you do not know what v is at the point of interest.
Please try using \dfrac{E}{2}=PE and post your steps.
 
  • Like
Likes   Reactions: bob tran
  • #15
<br /> k=20, m=0.5, v=1.5, x=?\\<br /> E=PE+KE\\<br /> E=\dfrac{E}{2}+KE\\<br /> E-\dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=\dfrac{1}{2}mv^2\\<br /> E=mv^2\\<br /> \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\<br /> \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\<br /> ...\\<br /> \texttt{which leads to}\\<br /> x=0.24 \ \texttt{m}<br />
Where am I going wrong?
Or if I just use \dfrac{E}{2}=PE itself:
<br /> \dfrac{E}{2}=PE\\<br /> \dfrac{1.125}{2}=\dfrac{1}{2}kx^2\\<br /> 1.125=kx^2\\<br /> \sqrt{\dfrac{1.125}{k}}=x\\<br /> \sqrt{\dfrac{1.125}{20}}=x\\<br /> 0.24 \ \texttt{m}=x<br />
Same answer..
 
Last edited:
  • #16
bob tran said:
<br /> k=20, m=0.5, v=1.5, x=?\\<br /> E=PE+KE\\<br /> E=\dfrac{E}{2}+KE\\<br /> E-\dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=\dfrac{1}{2}mv^2\\<br /> E=mv^2\\<br /> \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\<br /> \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\<br /> ...\\<br /> \texttt{which leads to}\\<br /> x=2.4 \ \texttt{m}<br />
Where am I going wrong?
All your algebra steps are correct. It is only when you plug in the values and solve for x that you somehow make a mistake. You should be getting 0.24 m, for some reason you are off by a factor of 10. Did you use the values given in your first post?
(By the way, there are actually two possible answers for x, 0.24 m and -0.24 m. I don't know why they choose to give only the positive answer)
 
  • Like
Likes   Reactions: bob tran
  • #17
nrqed said:
All your algebra steps are correct. It is only when you plug in the values and solve for x that you somehow make a mistake. You should be getting 0.24 m, for some reason you are off by a factor of 10. Did you use the values given in your first post?
(By the way, there are actually two possible answers for x, 0.24 m and -0.24 m. I don't know why they choose to give only the positive answer)
Sorry, it is 0.24 m. I just accidentally typed 2.4 m (fixed it in previous post now).

But, I am trying to get 0.17 m for part B, not 0.24 m.
 
  • #18
bob tran said:
<br /> k=20, m=0.5, v=1.5, x=?\\<br /> E=PE+KE\\<br /> E=\dfrac{E}{2}+KE\\<br /> E-\dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=\dfrac{1}{2}mv^2\\<br /> E=mv^2\\<br /> \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\<br /> \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\<br /> ...\\<br /> \texttt{which leads to}\\<br /> x=0.24 \ \texttt{m}<br />
Where am I going wrong?
Or if I just use \dfrac{E}{2}=PE itself:
<br /> \dfrac{E}{2}=PE\\<br /> \dfrac{1.125}{2}=\dfrac{1}{2}kx^2\\<br /> 1.125=kx^2\\<br /> \sqrt{\dfrac{1.125}{k}}=x\\<br /> \sqrt{\dfrac{1.125}{20}}=x\\<br /> 0.24 \ \texttt{m}=x<br />
Same answer..
bob tran said:
<br /> k=20, m=0.5, v=1.5, x=?\\<br /> E=PE+KE\\<br /> E=\dfrac{E}{2}+KE\\<br /> E-\dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=\dfrac{1}{2}mv^2\\<br /> E=mv^2\\<br /> \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\<br /> \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\<br /> ...\\<br /> \texttt{which leads to}\\<br /> x=0.24 \ \texttt{m}<br />
Where am I going wrong?
Or if I just use \dfrac{E}{2}=PE itself:
<br /> \dfrac{E}{2}=PE\\<br /> \dfrac{1.125}{2}=\dfrac{1}{2}kx^2\\<br /> 1.125=kx^2\\<br /> \sqrt{\dfrac{1.125}{k}}=x\\<br /> \sqrt{\dfrac{1.125}{20}}=x\\<br /> 0.24 \ \texttt{m}=x<br />
Same answer..
bob tran said:
Still not sure how I use this for the problem. :frown:

\dfrac{E}{2}=PE? But, what exactly am I supposed to do with that? When I substitute this into the equation, it still get 0.24 m as the answer. It seems like I don't understand something fundamental about this or I keep substituting in the wrong thing.
bob tran said:
Sorry, it is 0.24 m. I just accidentally typed 2.4 m (fixed it in previous post now).

But, I am trying to get 0.17 m for part B, not 0.24 m.
In your first post you used max KE = max PE to calculate the amplitude...great...you are more than 1/2 way there
Max PE = 1/2kA^2... PE at any other value of displacement, x is = 1/2kx^2
What value of x gives you 1/2 the max energy?...
 
  • Like
Likes   Reactions: bob tran
  • #19
nrqed said:
All your algebra steps are correct. It is only when you plug in the values and solve for x that you somehow make a mistake. You should be getting 0.24 m, for some reason you are off by a factor of 10. Did you use the values given in your first post?
(By the way, there are actually two possible answers for x, 0.24 m and -0.24 m. I don't know why they choose to give only the positive answer)
Ah, sorry I had misread the first post.

I see. The problem here is that in your solution, the v should be the speed when the object has PE = KE, but this speed is not 1.5 m/s (that's the maximum speed). Yours steps are correct, but using v=1/5 m/s in the final step is the mistake. When writing KE and PE, on must keep track where these are evaluated.

What you do know is that E = KE + PE and that at the equilibrium position, where PE =0, the speed is 1.5 m/s You therefore have a value for the total energy.

When you write E/2 = KE, this is correct but keep in mind that on the right side, the speed in 1/2 mv^2 is the speed at the position where KE=PE so this is an unknown. You can use that equation to find the speed at that point if you want, although it is not asked so you don't have to.

It is better to write E = PE + KE as E = PE + E/2 at the point of interest, and then solve for x.
 
  • Like
Likes   Reactions: bob tran
  • #20
nrqed said:
Ah, sorry I had misread the first post.

I see. The problem here is that in your solution, the v should be the speed when the object has PE = KE, but this speed is not 1.5 m/s (that's the maximum speed). Yours steps are correct, but using v=1/5 m/s in the final step is the mistake. When writing KE and PE, on must keep track where these are evaluated.

What you do know is that E = KE + PE and that at the equilibrium position, where PE =0, the speed is 1.5 m/s You therefore have a value for the total energy.

When you write E/2 = KE, this is correct but keep in mind that on the right side, the speed in 1/2 mv^2 is the speed at the position where KE=PE so this is an unknown. You can use that equation to find the speed at that point if you want, although it is not asked so you don't have to.

It is better to write E = PE + KE as E = PE + E/2 at the point of interest, and then solve for x.
Okay, hopefully, I got it this time.
<br /> \text{At the equilibrium position, PE=0, so:}\\<br /> E=\dfrac{1}{2}mv^2=\dfrac{1}{2}(.5)(1.5)^2=.5625 \text{ J}\\<br /> \text{So, because } \dfrac{E}{2}=KE,\\<br /> E=PE+\dfrac{E}{2}\\<br /> .5625=\dfrac{1}{2}kx^2+\dfrac{.5625}{2}\\<br /> \sqrt{\dfrac{.5625}{k}}=x\\<br /> \sqrt{\dfrac{.5625}{20}}=x\\<br /> .167750983=x\\<br /> 0.17 \text{ m}\approx x<br />
Is this the correct way of doing part B?
 
  • #21
bob tran said:
Okay, hopefully, I got it this time.
<br /> \text{At } PE=KE,\\<br /> E=\dfrac{1}{2}mv^2=\dfrac{1}{2}(.5)(1.5)^2=.5625 \text{ J}\\<br /> \text{So, because } \dfrac{E}{2}=KE,\\<br /> E=PE+\dfrac{E}{2}\\<br /> .5625=\dfrac{1}{2}kx^2+\dfrac{.5625}{2}\\<br /> \sqrt{\dfrac{.28125}{k}}=x\\<br /> \sqrt{\dfrac{.28125}{20}}=x\\<br /> .167750983=x\\<br /> 0.17 \text{ m}\approx x<br />
Is this the correct way of doing part B?
Yes, that's perfect.
What is important is to keep track (if not on paper at least in your head) of where the KE and PE are evaluated whenever you write a relation. For example, when you write E/2=KE , you have to remember that this true only at the point where KE = PE.

But yes, your derivation is correct.
 
  • Like
Likes   Reactions: bob tran
  • #22
bob tran said:
<br /> k=20, m=0.5, v=1.5, x=?\\<br /> E=PE+KE\\<br /> E=\dfrac{E}{2}+KE\\<br /> E-\dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=KE\\<br /> \dfrac{E}{2}=\dfrac{1}{2}mv^2\\<br /> E=mv^2\\<br /> \dfrac{1}{2}kx^2+\dfrac{1}{2}mv^2=mv^2\\<br /> \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\<br /> ...\\<br /> \texttt{which leads to}\\<br /> x=0.24 \ \texttt{m}<br />
Where am I going wrong?
Or if I just use \dfrac{E}{2}=PE itself:
<br /> \dfrac{E}{2}=PE\\<br /> \dfrac{1.125}{2}=\dfrac{1}{2}kx^2\\<br /> 1.125=kx^2\\<br /> \sqrt{\dfrac{1.125}{k}}=x\\<br /> \sqrt{\dfrac{1.125}{20}}=x\\<br /> 0.24 \ \texttt{m}=x<br />
Same answer..
E is not 1.125J. How did you get that? It should be half that.
 
  • #23
lychette said:
In your first post you used max KE = max PE to calculate the amplitude...great...you are more than 1/2 way there
Max PE = 1/2kA^2... PE at any other value of displacement, x is = 1/2kx^2
What value of x gives you 1/2 the max energy?...

lychette said:
In your first post you used max KE = max PE to calculate the amplitude...great...you are more than 1/2 way there
Max PE = 1/2kA^2... PE at any other value of displacement, x is = 1/2kx^2
What value of x gives you 1/2 the max energy?...[
Thank you...you had the answer in reach in your first post...you were diverted from the straight and narrow and got lost in the woods :)
 

Similar threads

An exploding cannon shell
  • · Replies 4 ·
Replies
4
Views
1K
Amplitude and Period of Oscillation from Collision
  • · Replies 2 ·
Replies
2
Views
5K
Sliding block hits and compresses a spring
  • · Replies 5 ·
Replies
5
Views
1K
Oscillations of a disc with a smaller disc removed (Feynman ex. 17.23)
  • · Replies 2 ·
Replies
2
Views
2K
Using energy considerations to analyse particle motion
  • · Replies 3 ·
Replies
3
Views
1K
Differential Equation for a Pendulum
  • · Replies 21 ·
Replies
21
Views
3K
Finding Spring Constant & Energy w/ Doubt in Exercise
  • · Replies 7 ·
Replies
7
Views
2K
Ranking Kinetic Energy: Why Does Mass Matter More Than Amplitude?
  • · Replies 22 ·
Replies
22
Views
4K
Elastic potential energy - different methods, different results
  • · Replies 2 ·
Replies
2
Views
1K
Does amplitude depend on mass in SHM?
  • · Replies 1 ·
Replies
1
Views
8K