Pan suspended by a spring (Energy + SHM)

[uSee2]

Homework Statement

An ideal massless spring is hung from the ceiling and a pan suspension of total mass M is suspended from the end of
the spring. A piece of clay, also of mass M, is then dropped from a height H onto the pan and sticks to it.
Express all
algebraic answers in terms of the given quantities and fundamental constants.
(a) Determine the speed of the clay at the instant it hits the pan.
(b) Determine the speed of the clay and pan just after the clay strikes it.
(c) After the collision, the apparatus comes to rest at a distance H/2 below the current position. Determine the
spring constant of the attached spring. (d) Determine the resulting period of oscillation

Relevant Equations

$U_g = mgh$
$KE = 0.5mv^2$

I have successfully completed parts A, and B, however, I am confused on Part C. Here was my attempt and the answer key's attempt:

My attempt:
Since I correctly knew the speed after the collision, and the gravitational potential energy after the collision if I set h=0 at when it was at rest temporarily, then the change in spring potential energy is equal to the sum of the KE and Gravitational PE combined. Since when it is at rest at the bottom is has none of those except spring Potential Energy.

As such:
$KE + PE_g = \Delta PE_s$
This is where my attempt and key's attempt starts to branch off.

They stated that $\Delta PE_s = \frac 1 2 k(\frac H 2)^2$

I stated that $\Delta PE_s = \frac 1 2 kHx_i + \frac 1 8 kH^2$ (Since if calculated manually, that is what $\Delta PE_s$ is equal to, I did my calculating below)
where $x_i$ is initial stretch before the collision to account for the initial stretch of the spring when the pan was just hanging on it before the collision.

They specifically stated the reason why they ignored the initial stretch of the spring was:

"Even though the initial current rest position immediately after the collision has an unknown initial stretch to begin with due to the weight of the tray and contains spring energy, we can set this as the zero spring energy position and use the additional stretch distance H/2 given to equate the conversion of kinetic and gravitational energy after the collision into the additional spring energy gained at the end of stretch. "

However I don't think that this is true, since if calculated using
$\Delta PE_s = PE_{sf} - PE_{si}$
Where:
$PE_{sf} = \frac 1 2 k(x_i + \frac H 2)^2$ Spring PE finally
$PE_{si} = \frac 1 2 kx_i^2$ Spring PE initially
it does not yield:
$\Delta PE_s = \frac 1 2 k(\frac H 2)^2$
If multiplied out the correct change in $PE_s$ is:
$\Delta PE_s = \frac 1 2 kHx_i + \frac 1 8 kH^2$

My question is, why could the key ignore the initial spring potential energy and set it at the "zero" point? Wouldn't the initial stretch need to be somehow accounted for?

 

[haruspex]

Homework Statement:: An ideal massless spring is hung from the ceiling and a pan suspension of total mass M is suspended from the end of
the spring. A piece of clay, also of mass M, is then dropped from a height H onto the pan and sticks to it.
Express all
algebraic answers in terms of the given quantities and fundamental constants.
(a) Determine the speed of the clay at the instant it hits the pan.
(b) Determine the speed of the clay and pan just after the clay strikes it.
(c) After the collision, the apparatus comes to rest at a distance H/2 below the current position. Determine the
spring constant of the attached spring. (d) Determine the resulting period of oscillation
Relevant Equations:: $U_g = mgh$
$KE = 0.5mv^2$

View attachment 320165
I have successfully completed parts A, and B, however, I am confused on Part C. Here was my attempt and the answer key's attempt:

My attempt:
Since I correctly knew the speed after the collision, and the gravitational potential energy after the collision if I set h=0 at when it was at rest temporarily, then the change in spring potential energy is equal to the sum of the KE and Gravitational PE combined. Since when it is at rest at the bottom is has none of those except spring Potential Energy.

As such:
$KE + PE_g = \Delta PE_s$
This is where my attempt and key's attempt starts to branch off.

They stated that $\Delta PE_s = \frac 1 2 k(\frac H 2)^2$

I stated that $\Delta PE_s = \frac 1 2 kHx_i + \frac 1 8 kH^2$ (Since if calculated manually, that is what $\Delta PE_s$ is equal to, I did my calculating below)
where $x_i$ is initial stretch before the collision to account for the initial stretch of the spring when the pan was just hanging on it before the collision.

They specifically stated the reason why they ignored the initial stretch of the spring was:

"Even though the initial current rest position immediately after the collision has an unknown initial stretch to begin with due to the weight of the tray and contains spring energy, we can set this as the zero spring energy position and use the additional stretch distance H/2 given to equate the conversion of kinetic and gravitational energy after the collision into the additional spring energy gained at the end of stretch. "

However I don't think that this is true, since if calculated using
$\Delta PE_s = PE_{sf} - PE_{si}$
Where:
$PE_{sf} = \frac 1 2 k(x_i + \frac H 2)^2$ Spring PE finally
$PE_{si} = \frac 1 2 kx_i^2$ Spring PE initially
it does not yield:
$\Delta PE_s = \frac 1 2 k(\frac H 2)^2$
If multiplied out the correct change in $PE_s$ is:
$\Delta PE_s = \frac 1 2 kHx_i + \frac 1 8 kH^2$

My question is, why could the key ignore the initial spring potential energy and set it at the "zero" point? Wouldn't the initial stretch need to be somehow accounted for?

You are quite correct. Where does this question come from?
Edit…
I have found it, almost: . But part c is now a different question.

 

[kuruman]

My question is, why could the key ignore the initial spring potential energy and set it at the "zero" point? Wouldn't the initial stretch need to be somehow accounted for?

Actually, the initial stretch is taken into account but not how you think. At equilibrium, the vertical spring is elongated from its relaxed position by $\Delta x = mg/k$. The choice of the zero of any potential energy is arbitrary. In this case, if the zero of elastic potential energy is chosen at this equilibrium position and $\xi$ is a parameter defining the spring's displacement from this equilibrium position, then any gravitational potential energy change is automatically taken into account. The vertical spring-mass system becomes equivalent to a horizontal where $\xi(t)$ is the displacement of the oscillating mass relative to the vertical equilibrium position.

The statement

"Even though the initial current rest position immediately after the collision has an unknown initial stretch to begin with due to the weight of the tray and contains spring energy, we can set this as the zero spring energy position and use the additional stretch distance H/2 given to equate the conversion of kinetic and gravitational energy after the collision into the additional spring energy gained at the end of stretch. "

is appropriate. You can find a mathematical derivation of why here.

 

[haruspex]

Actually, the initial stretch is taken into account but not how you think. At equilibrium, the vertical spring is elongated from its relaxed position by $\Delta x = mg/k$. The choice of the zero of any potential energy is arbitrary. In this case, if the zero of elastic potential energy is chosen at this equilibrium position and $\xi$ is a parameter defining the spring's displacement from this equilibrium position, then any gravitational potential energy change is automatically taken into account. The vertical spring-mass system becomes equivalent to a horizontal where $\xi(t)$ is the displacement of the oscillating mass relative to the vertical equilibrium position.

The statement

is appropriate. You can find a mathematical derivation of why here.

Ok, but that works by redefining the elastic PE to include the GPE. In effect, the extra tension (the tension at equilibrium) is taken as neutralising gravity at all positions.
@uSee2 quotes the source

They stated that $\Delta PE_s = \frac 1 2 k(\frac H 2)^2$

which is only true after such a redefinition. Perhaps we need to see the whole of the source, but at the least it sounds misleading.

Edit… the other problem is that the mass that caused the initial equilibrium extension is only M, while the mass that subsequently descends is 2M. So the lost GPE is double the ignored EPE.

 

[uSee2]

Ok, but that works by redefining the elastic PE to include the GPE. In effect, the extra tension (the tension at equilibrium) is taken as neutralising gravity at all positions.
@uSee2 quotes the source

which is only true after such a redefinition. Perhaps we need to see the whole of the source, but at the least it sounds misleading.

I believe that this was the original source (Page 15) and the answer is on page 26.
https://www.npsd.k12.nj.us/cms/lib04/NJ01001216/Centricity/Domain/473/Oscillations and SHM.pdf

 

[haruspex]

I believe that this was the original source (Page 15) and the answer is on page 26.
https://www.npsd.k12.nj.us/cms/lib04/NJ01001216/Centricity/Domain/473/Oscillations and SHM.pdf

I see that gives the answer as $k=\frac{12Mg}H$, which is not what I get.
What do you get?
I have notified npsdnj of their error.

 

[uSee2]

I see that gives the answer as $k=\frac{12Mg}H$, which is not what I get.
What do you get?
I have notified npsdnj of their error.

Thank you! I was not able to get an acceptable answer since I could not remove initial spring stretch. Do you think that this problem is impossible without the initial stretch length?

 

[haruspex]

Thank you! I was not able to get an acceptable answer since I could not remove initial spring stretch. Do you think that this problem is impossible without the initial stretch length?

You can express that initial length in terms of k, M and g. A solution can be obtained.

 

[kuruman]

I see that gives the answer as $k=\frac{12Mg}{H}$, which is not what I get.
What do you get?
I have notified npsdnj of their error.

I got $k=\dfrac{12Mg}{h}$. I will send you my solution via PM for you to verify.

 

[kuruman]

I see that gives the answer as $k=\frac{12Mg}H$, which is not what I get.
What do you get?
I have notified npsdnj of their error.

I think I found a mistake in my solution when I started putting it in LaTeX. I will course send you my final answer and explain the mistake I made which gave me their answer.

 

[uSee2]

You can express that initial length in terms of k, M and g. A solution can be obtained.

That is right, I forgot about that!

$\Delta PE_s = \frac 1 2 kHx_i + \frac 1 8 kH^2$ from above
$kx_i = Mg$ Before collision when it was just hanging
$x_i = \frac {Mg} {k}$
$\Delta PE_s = \frac 1 2 kH\frac {Mg} {k} + \frac 1 8 kH^2$
K's cancel
$\Delta PE_s = \frac 1 2 HMg + \frac 1 8 kH^2$
$KE = \frac 1 2 (2M) (v_f)^2$
$PE_g = (2M)g(\frac H 2)$
Since $v_f = \frac 1 2 \sqrt{gH}$ from conservation of momentum from parts B & C
$KE = \frac 1 4 MgH$
$PE_g = MgH$
$KE + PE_g = \frac 5 4 MgH = \Delta PE_s$
$\frac 1 2 HMg + \frac 1 8 kH^2 = \frac 5 4 MgH$
$\frac 1 8 kH^2 = \frac 9 8 MgH$
Solving further for k yields:
$k = \frac {9Mg} {H}$
Is this correct?

 

[haruspex]

Since $v_f = \frac 1 2 \sqrt{gH}$ from conservation of momentum from parts B & C

Is that really what you got for part b?

 

[Chestermiller]

I'm never comfortable solving a problem like this using only energy balance. I like to start with force balances and then integrate these as necessary to arrive at the energy balance.

Let $x_0$ represent the undeformed length of the spring and $x_i$ be the length of the spring with just the pan of mass M hanging from it. Then $$k(x_i-x_0)=Mg\tag{1}$$The velocity of the falling mass M when it hits the pan is $v=\sqrt{2gH}$, so, from a momentum balance, the downward velocity of the mass and pan immediately after collision is $$\frac{dx}{dt}=v_i=\sqrt{\frac{gH}{2}}$$A downward force balance on the masses subsequent to the collision will read: $$2Mg-k(x-x_0)=2Mv\frac{dv}{dx}\tag{2}$$Adding Eqns. 1 and 2 then yields: $$Mg-k(x-x_i)=2Mv\frac{dv}{dx}\tag{3}$$What do you get if you ntegrate this equation between $x=x_i$ and $x=x_i+\frac{H}{2}$?

 

[Chestermiller]

Hey, Chet. The OP is still working on this. Too soon for a full solution.

I changed it to allow the OP to integrate the force balance to obtain the energy balance. I hope this helps.

 

[haruspex]

I changed it to allow the OP to integrate the force balance to obtain the energy balance. I hope this helps.

@uSee2 almost had it in post #11. Looks like s/he went wrong either in solving part b or in miscopying that result when solving part c.

 

[uSee2]

Is that really what you got for part b?

I see where I went wrong, I accidently substituted in H/2 instead of H for my first kinematic equation. Speed after they collide is:

$\sqrt{2gH} = v_f$
New KE is:
$\frac 1 2 2Mv_f^2 = 2MgH = KE$
Doing all my calculations again is:
$\Delta PE_s = KE + PE_g$
$\frac 1 2 MgH + \frac 1 8 kH^2 = KE + PE_g$
$\frac 1 2 MgH + \frac 1 8 kH^2 = 2MgH+ MgH$
$\frac 1 2 MgH + \frac 1 8 kH^2 = 3MgH$
$\frac 1 8 kH^2 = 2.5MgH$
$k = \frac {20Mg}{H}$
Is this correct?

 

[haruspex]

I see where I went wrong, I accidently substituted in H/2 instead of H for my first kinematic equation. Speed after they collide is:

$\sqrt{2gH} = v_f$
New KE is:
$\frac 1 2 2Mv_f^2 = 2MgH = KE$
Doing all my calculations again is:
$\Delta PE_s = KE + PE_g$
$\frac 1 2 MgH + \frac 1 8 kH^2 = KE + PE_g$
$\frac 1 2 MgH + \frac 1 8 kH^2 = 2MgH+ MgH$
$\frac 1 2 MgH + \frac 1 8 kH^2 = 3MgH$
$\frac 1 8 kH^2 = 2.5MgH$
$k = \frac {20Mg}{H}$
Is this correct?

Sadly, no. $\sqrt{2gH}$ was the speed of the mass just before it hit the pan. What did you get for part b, the speed just after collision?

 

[uSee2]

Sadly, no. $\sqrt{2gH}$ was the speed of the mass just before it hit the pan. What did you get for part b, the speed just after collision?

Ahh I see, I forgot the mass doubles. The corrected speed is $\frac 1 2 \sqrt{2gH}$

After doing it all again, I got
$\frac {8Mg} {H} = k$ Is this correct?

 

[haruspex]

Ahh I see, I forgot the mass doubles. The corrected speed is $\frac 1 2 \sqrt{2gH}$

After doing it all again, I got
$\frac {8Mg} {H} = k$ Is this correct?

Yes!

 

[uSee2]

Yes!

Thank you so much! And so this is the correct answer rather than the $\frac {12Mg} {H}$ in the answer key?

 

[haruspex]

Thank you so much! And so this is the correct answer rather than the $\frac {12Mg} {H}$ in the answer key?

Yes.