Ranking Kinetic Energy: Why Does Mass Matter More Than Amplitude?

[Huski]

Homework Statement

Rank the cases from maximum Kinetic Energy (KE) to lowest KE. State any ties. Briefly, explain your ranking.

Case 1 has mass M and amplitude A
Case 2 has mass 2M and amplitude A
Case 3 has mass M and amplitude 2A

Homework Equations

1. $E=\dfrac{1}{2}kA^{2}$
2. $E=\dfrac{1}{2}mv^{2}$

k = spring constant (N/m) Newton/meter
A = amplitude (m) meter
v = max velocity (m/s) meters/second

The Attempt at a Solution

I have the answer and it's 3 > (1 = 2)
Now, I've been trying to understand why? Equation 1 includes amplitude and equation 2 includes mass. All 3 cases include mass and amplitude. In equation 1, as amplitude increases, kinetic energy increases. In equation 2, as mass increases (by slower margins because of the half in front), kinetic energy increases. I don't get why case 1 = case 2? They both have the same amplitude, but case 2 has more mass than case 1. I think case 2 should have more kinetic energy than case one. Here's why, if I look at an example below.

let's say mass equals 4. (note: I'm ignoring velocity)

Case 1: E = 0.5(4) = 2 Joules.
Case 2: E = 0.5(2*4) = 4 Joules.

So mass is relevant when ranking the kinetic energy. Can someone shed light on something I may be missing? Thank you.

 

[QuantumQuest]

In all three cases you have the potential energy of the spring i.e. $E=\dfrac{1}{2}kA^{2}$ that converts to kinetic energy of the mass i.e. $E=\dfrac{1}{2}mv^{2}$ Now, does applying a different mass change the potential energy of the spring? Consider this for the tie between cases 1 and 2. For case 3 does the mass get higher energy from the spring or not?

 

[Huski]

Now, does applying a different mass change the potential energy of the spring?

I would say for the first part, that mass doesn't change potential energy because PE depends on spring constant (k) + amplitude (A). Okay, so I first look at PE, then I see if the mass is there and if not (and it's obviously not), then when converting from PE to KE (because the spring begins moving) it doesn't' change the KE? Am I understanding this correctly?

For case 3 does the mass get higher energy from the spring or not?

For case 3, I would say that the spring constant doesn't increase the energy of the mass. It does in case 2, the spring would need more k value (force) to push more mass (2M). Not for case 3, I believe.

 

[QuantumQuest]

Am I understanding this correctly?

The potential energy of the spring between cases 1 and 2 is the same. What has changed in case 2 is the mass of the body. So, when you release the body in this case what does this mean for the velocity of the body taking into account that mass has doubled?

For case 3, I would say that the spring constant doesn't increase the energy of the mass. It does in case 2, the spring would need more k value (force) to push more mass (2M). Not for case 3, I believe.

In case 3 the amplitude is $2A$ as shown in the diagram. So, use the equation for the potential energy of the spring. Is the amount of higher energy transferred to the body higher or not?

 

[Huski]

what does this mean for the velocity of the body taking into account that mass has doubled?

So, I think that if KE stays the same, mass doubles, then velocity is half. I think this because I look at this equation below.

$E=\dfrac{1}{2}mv^{2}$

If I substitute 2 into the equation:

$E=\dfrac{1}{2}(2m)(v)^{2}$

$2 \cdot E=\dfrac{1}{2}mv^{2}$ Now I need to get 2 KE back to 1 KE, so I manipulate V

$2 \cdot E=\dfrac{1}{2}(m)(\dfrac{1}{2}v)^{2}$ then i multiply by reciprocal or divide by 1/2 to and get:

$E=\dfrac{1}{2}mv^{2}$

So if I double mass, I take half velocity.

Is the amount of higher transferred to the body higher or not?

Okay, so for higher energy transfer, doubling A (2A) will quadruple PE I believe. Then K needs to go up Edit: (if I wanted to keep energy the same). So energy is higher.

$E=\dfrac{1}{2}kA^{2}$

$E=\dfrac{1}{2}k(2A)^{2}$

$4 \cdot E=\dfrac{1}{2}kA^{2}$

$4 \cdot E=\dfrac{1}{2}(8k)A^{2}$

$4 \cdot E=(4k)A^{2}$

$E=\dfrac{1}{2}kA^{2}$

 

[QuantumQuest]

If I substitute 2 into the equation:
$E=\dfrac{1}{2}(2m)(v)^{2}$
$2 \cdot E=\dfrac{1}{2}mv^{2}$

Be careful with the math in the last equation in quote.

Okay, so for higher energy transfer, doubling A (2A) will quadruple PE I believe. Then K needs to go up Edit: (if I wanted to keep energy the same). So energy is higher.
$E=\dfrac{1}{2}kA^{2}$
$E=\dfrac{1}{2}k(2A)^{2}$
$4 \cdot E=\dfrac{1}{2}kA^{2}$

Be careful with the math in the last equation in quote.
I had a typo in my post #4

Is the amount of higher energy transferred to the body higher or not?

(I have struck through the wrong one).

Now, what do you mean by

Then K needs to go up

$k$ is a constant of the spring. So, what changes the potential energy is amplitude $A$.

 

[Huski]

I was just saying that k goes up if I wanted to cancel out the 4 on the left side of the equation. But because we don't care about that (because as you said k is constant) I made the "Edit" in there to explain why I said that.
So when you asked about mass doubling, velocity is half? And the amount of energy transferred is higher because of the amplitude (as you said).
So to clear things up, case 1 = case 2 because when converting from PE to KE, mass is irrelevant because what causes the energy to increase is the amplitude. More potential energy means more kinetic energy?

 

[QuantumQuest]

I was just saying that k goes up if I wanted to cancel out the 4 on the left side of the equation. But because we don't care about that (because as you said k is constant) I made the "Edit" in there to explain why I said that.
So when you asked about mass doubling, velocity is half?

I would suggest to do the math correctly for the two cases I talked about in my post #6. This is crucial in order to see how things work.

And the amount of energy transferred is higher because of the amplitude (as you said).

Yes

So to clear things up, case 1 = case 2 because when converting from PE to KE, mass is irrelevant because what causes the energy to increase is the amplitude. More potential energy means more kinetic energy?

For both cases 1 and 2, the amount of energy transferred from spring to body is the same. So, what changes is the mass of the body (doubles) and the velocity of the body decreases such that the $E$ in $E=\dfrac{1}{2}mv^{2}$ stays the same. Just do the math to see what is the final expression. Now, more potential energy from spring means that more energy will be transferred to the body so if you release it, it will have more kinetic energy.

 

[Huski]

So you said that I did the last part of the equation wrong : $2\cdot E=\dfrac{1}{2}mv^{2}$.

is this correct?$E=\dfrac{1}{2}(2m)v^2$ 1/2 cancels out 2

$E=mv^2$

 

[QuantumQuest]

is this correct?$E=\dfrac{1}{2}(2m)v^2$ 1/2 cancels out 2

$E=mv^2$

Yes, that's correct. Also take care of $v$. This is not the same in case 2 compared to case 1.

 

[Huski]

Okay so,

case 2: $E=mv^2$

case 1: $E=\dfrac{1}{2}(m)v^2$

 

[haruspex]

if KE stays the same, mass doubles, then velocity is half

Not quite right. Take another look at the equation.

 

[haruspex]

Okay so,

case 2: $E=mv^2$

case 1: $E=\dfrac{1}{2}(m)v^2$

As quantumquest wrote, distinguish the velocities. They are not the same.

 

[Huski]

Not quite right. Take another look at the equation.

So if KE stays the same, mass doubles, then velocity stays the same?

I think this might be it:

case 2: $E=mv^2$

case 1: $E=\dfrac{1}{2}(m)(2v)^2$

 

[haruspex]

So if KE stays the same, mass doubles, then velocity stays the same?

No, not that either.
The equation is E=½mv². If you double the mass with the same KE then clearly v must change, but by how much? Look at what happens if we try halving v:
½(2m)(v/2)²=?

 

[Huski]

Well, I think that if you double the velocity, you increase the KE by a factor of 4. So I think you need velocity to change by 1/4?
case 1: $E=\dfrac{1}{2}(m)\dfrac{1}{4}v^2$

 

[QuantumQuest]

I think this might be it:

case 2: $E=mv^2$

case 1: $E=\dfrac{1}{2}(m)(2v)^2$

Also, in addition to what haruspex said, it may be of help to give a different name for the velocity in case 2 e.g. $v_1$. Then you'll have just to equate the kinetic energies of case 1 and 2 i.e $\frac{1}{2}mv^2$ and $\frac{1}{2}(2m)v_1^2$ and do the math.

 

[haruspex]

Well, I think that if you double the velocity, you increase the KE by a factor of 4. So I think you need velocity to change by 1/4?
case 1: $E=\dfrac{1}{2}(m)\dfrac{1}{4}v^2$

Please, no more wild guesses. Follow Qq's advice in post #17 and use algebra to find the answer.

 

[Huski]

This is what I came up with: $E=\dfrac{1}{2}(2\cdot m)(0.707\cdot v_{1})^2$

 

[haruspex]

This is what I came up with: $E=\dfrac{1}{2}(2\cdot m)(0.707\cdot v_{1})^2$

Right, the speed is reduced by a factor √2.

 

[Huski]

So we found the scale factors for the variables, but the kinetic energy omits the mass and we care for the amplitude?

 

[haruspex]

the kinetic energy omits the mass

No, the KE involves mass and velocity, but the problem is that changing the mass while keeping k constant changed the velocity, so we cannot use the usual KE equation to find the energy.
Instead, we use the fact that the total energy is constant in SHM. This means that that the peak KE (i.e. at force equilibrium) is the same as peak PE (i.e. when speed is zero). So we can use the PE equation to find the peak KE.

 

[Huski]

This makes sense. thank you, guys!