Simple Harmonic Motion - Potential energy

[PFuser1232]

For a mass on a spring (vertical set up) why is potential energy U defined as 1/2 kx^2? This is just the elastic potential energy. Shouldn't it be U = 1/2 kx^2 + mgh? Both the elastic AND potential energy? Also, for a simple pendulum at a very low amplitude, the potential energy is all gravitational, right? By the way, if U = 1/2 kx^2 + mgh, then the relationship ω = √(k/m) becomes invalid, since U = 1/2 mω^2 x^2 is always valid and equating this to both elastic and gravitational potential energy gives us a different expression for ω.

 

[nasu]

1/2 kx^2 is the only term that is not constant in the potential energy.
The gravitational potential energy can add a constant to it. As you can always do for potential energy.
Try to calculate is with gravity included and you will see it.

 

[olivermsun]

Hint: Define x so that x=0 right where the system is at "rest." (This will not be the same position as the rest position in the absence of gravity.)

 

[russ_watters]

Consider that if your rotate the system horizontal (and put it on a frictionless surface), the equations still work. Gravity only determines the equilibrium position, it doesn't determine the dynamics.

 

[PFuser1232]

Consider that if your rotate the system horizontal (and put it on a frictionless surface), the equations still work. Gravity only determines the equilibrium position, it doesn't determine the dynamics.

I know it doesn't determine the dynamics, but this is not my doubt. My doubt is, 1/2 m w^2 x^2 must always equal 1/2 kx^2 yet this is not the case here, because there is gravitational potential energy varying as the mass oscillates up and down. The conservation of energy holds whether or not gravity affects the dynamics of the mass on the spring.

 

[PFuser1232]

1/2 kx^2 is the only term that is not constant in the potential energy.
The gravitational potential energy can add a constant to it. As you can always do for potential energy.
Try to calculate is with gravity included and you will see it.

How is gravitational potential energy constant? The mass is moving up and down. And K_A + U_totA = K_B + U_totB always holds as mechanical energy is conserved, where U_tot = U_elastic + U_grav. Can you elaborate more please?

 

[jtbell]

First suppose there is no gravity, and we have the spring oriented vertically with one end fixed. Let x = 0 be the position of the mass when it is in equilibrium. Then
$$U_{spring} = \frac{1}{2}kx^2$$
Next, remove the spring for a moment, and "turn on" gravity. Define the gravitational potential energy such that it is zero when the mass is at the same x = 0 position as defined above. Then
$$U_{grav} = mgx$$

Under the influence of both the spring and gravity, we therefore have
$$U = \frac{1}{2}kx^2 + mgx$$
In this situation the mass has a new equilibrium position, where $\vec F_{spring} + \vec F_{grav} = 0$.

Exercise 1: Find the new equilibrium position... how far is it below the old equilibrium position?

Exercise 2: Define a new vertical position variable, z, such that z = 0 at the new equilibrium position. Find an equation that relates z and x.

Exercise 3: Change the variable in the equation for U above, from x to z, by making an appropriate substitution. What do you get?

 

[olivermsun]

The point is that when you dangle the mass under its own weight, the equilibrium position moves from the point determined by the rest length of the spring (call it x = 0) to some new lower position (x = c < 0) where F = -kc = -mg. That is, the new rest position is exactly where the stretched spring exerts a restoring force that exactly opposes gravity.

Due to the linearity of the governing equation, you can analyze the system as:

F = -kx + mg = -kx + kc = -k(x-c) .

So if you consider this as a system in the new variable x' = x - c, then you see that everything behaves exactly as before, except now the center of oscillation is around x' = 0 (where x = c).

(Never mind, jtbell already posted the solution...)

 

[nasu]

How is gravitational potential energy constant? The mass is moving up and down. And K_A + U_totA = K_B + U_totB always holds as mechanical energy is conserved, where U_tot = U_elastic + U_grav. Can you elaborate more please?

I did not say that the gravitational potential energy is constant, did I?
I said that considering the gravitational potential energy results in adding a constant to the expression of the energy
After you do the work, as suggested already by jtbel.
The best way to understand is to try yourself and do the calculations.

 

[PFuser1232]

First suppose there is no gravity, and we have the spring oriented vertically with one end fixed. Let x = 0 be the position of the mass when it is in equilibrium. Then
$$U_{spring} = \frac{1}{2}kx^2$$
Next, remove the spring for a moment, and "turn on" gravity. Define the gravitational potential energy such that it is zero when the mass is at the same x = 0 position as defined above. Then
$$U_{grav} = mgx$$

Under the influence of both the spring and gravity, we therefore have
$$U = \frac{1}{2}kx^2 + mgx$$
In this situation the mass has a new equilibrium position, where $\vec F_{spring} + \vec F_{grav} = 0$.

Exercise 1: Find the new equilibrium position... how far is it below the old equilibrium position?

Exercise 2: Define a new vertical position variable, z, such that z = 0 at the new equilibrium position. Find an equation that relates z and x.

Exercise 3: Change the variable in the equation for U above, from x to z, by making an appropriate substitution. What do you get?

I got lost at Ex 2, can you please give me more hints?

 

[PFuser1232]

The point is that when you dangle the mass under its own weight, the equilibrium position moves from the point determined by the rest length of the spring (call it x = 0) to some new lower position (x = c < 0) where F = -kc = -mg. That is, the new rest position is exactly where the stretched spring exerts a restoring force that exactly opposes gravity.

Due to the linearity of the governing equation, you can analyze the system as:

F = -kx + mg = -kx + kc = -k(x-c) .

So if you consider this as a system in the new variable x' = x - c, then you see that everything behaves exactly as before, except now the center of oscillation is around x' = 0 (where x = c).

(Never mind, jtbell already posted the solution...)

Do you mean to say that now the total potential energy is 1/2 k x' ^2 ?

 

[UltrafastPED]

For a nicely illustrated lecture see: http://www.ux1.eiu.edu/~cfadd/1150/15Period/Vert.html

 

[PFuser1232]

First suppose there is no gravity, and we have the spring oriented vertically with one end fixed. Let x = 0 be the position of the mass when it is in equilibrium. Then
$$U_{spring} = \frac{1}{2}kx^2$$
Next, remove the spring for a moment, and "turn on" gravity. Define the gravitational potential energy such that it is zero when the mass is at the same x = 0 position as defined above. Then
$$U_{grav} = mgx$$

Under the influence of both the spring and gravity, we therefore have
$$U = \frac{1}{2}kx^2 + mgx$$
In this situation the mass has a new equilibrium position, where $\vec F_{spring} + \vec F_{grav} = 0$.

Exercise 1: Find the new equilibrium position... how far is it below the old equilibrium position?

Exercise 2: Define a new vertical position variable, z, such that z = 0 at the new equilibrium position. Find an equation that relates z and x.

Exercise 3: Change the variable in the equation for U above, from x to z, by making an appropriate substitution. What do you get?

Was i supposed to get U = 1/2 kz^2? I ended up with a slightly more complicated answer.

 

[PFuser1232]

The point is that when you dangle the mass under its own weight, the equilibrium position moves from the point determined by the rest length of the spring (call it x = 0) to some new lower position (x = c < 0) where F = -kc = -mg. That is, the new rest position is exactly where the stretched spring exerts a restoring force that exactly opposes gravity.

Due to the linearity of the governing equation, you can analyze the system as:

F = -kx + mg = -kx + kc = -k(x-c) .

So if you consider this as a system in the new variable x' = x - c, then you see that everything behaves exactly as before, except now the center of oscillation is around x' = 0 (where x = c).

(Never mind, jtbell already posted the solution...)

Apparently I made a wrong substitution in the equation U = mgx + 1/2 kx^2. Can you please tell me where I went wrong?

 

[jtbell]

Was i supposed to get U = 1/2 kz^2?

Correct!

I ended up with a slightly more complicated answer.

Show us what you did and someone will probably be able to find your error. When you do the substitution you should get something a bit messy, but most of it should cancel out, leaving only the 1/2 kz^2.

 

[jtbell]

I got lost at Ex 2, can you please give me more hints?

Here's a diagram which might help. I don't know what you got for Ex 1, so I'm calling it Δ. For any vertical position, what is the relationship between z and x?

 

[PFuser1232]

Correct!
Show us what you did and someone will probably be able to find your error. When you do the substitution you should get something a bit messy, but most of it should cancel out, leaving only the 1/2 kz^2.

http://postimg.org/image/68566ta6f/

This is my working, where did I go wrong?

U = 1/2 kx^2 + mgx, and x = z + c, and mg = kc
I eventually got U = 1/2 kz^2 + 2kzc + 3/2 kc^2

 

[PFuser1232]

Correct!



Show us what you did and someone will probably be able to find your error. When you do the substitution you should get something a bit messy, but most of it should cancel out, leaving only the 1/2 kz^2.

When I tried to modify what I did by using mg = -kc instead, i ended up with U = 1/2k(z^2 - c^2)
Apparently I have a misconception regarding the signs. The thing is, throughout my mechanics courses, If I have an equilibrium problem in which, say, a mass is on a spring, I would equate T to mg where T is the tension and mg is the force of gravity, but I never did something like T = -mg, or anything of that sort. Where exactly am I going wrong?

 

[nasu]

It's a matter of choosing the positive direction.
You wrote the energy with +mgx for gravitational PE.
x is measured from the position of equilibrium without gravity, right? Where the spring is not stretched.
When the body goes down the gravitational PE must decrease. This means that x must be negative.
So you implicitly assumed that negative is down and positive is up and your "x" is negative.
This is why the condition of equilibrium is
mg=-kx.

You can avoid this by
Writing the gravitational PE term as -mgx.

 

[PFuser1232]

It's a matter of choosing the positive direction.
You wrote the energy with +mgx for gravitational PE.
x is measured from the position of equilibrium without gravity, right? Where the spring is not stretched.
When the body goes down the gravitational PE must decrease. This means that x must be negative.
So you implicitly assumed that negative is down and positive is up and your "x" is negative.
This is why the condition of equilibrium is
mg=-kx.

You can avoid this by
Writing the gravitational PE term as -mgx.

But that still doesn't give U = 1/2 kz^2

 

[PFuser1232]

Can someone please check my working and tell me where exactly I went wrong?

 

[nasu]

But that still doesn't give U = 1/2 kz^2

It does. Up to a constant term. This is what I told you.
Adding a constant to the potential energy does not make any difference.
The changes in PE are the same.

 

[PFuser1232]

It does. Up to a constant term. This is what I told you.
Adding a constant to the potential energy does not make any difference.
The changes in PE are the same.

http://postimg.org/image/xousm7yjt/ check this
First of all, is my expression for potential energy correct? If it is, then how is the relationship ω = √(k/m) satisfied?

 

[jtbell]

U = 1/2 kx^2 + mgx, and x = z + c, and mg = kc
I eventually got U = 1/2 kz^2 + 2kzc + 3/2 kc^2

x = z + c is not consistent with my diagram and my sign convention (which is that both x and z increase in the upwards direction). Looking at the first horizontal dashed line on my diagram, for example, if z = +2c, what do you have to do to it to get x = +c? (Your c is the same as my Δ).

But even after you fix that, U still ends up with a constant term along with the (1/2)kz^2. I must have made a mistake in my algebra yesterday (which I don't have here so I can't check it now). Sorry about that.

Nevertheless, that constant term is OK, because adding a constant to the potential energy everywhere doesn't change any physical results. What matters physically is the difference in potential energy, ΔU, between two different locations, and a constant term always cancels out when you calculate ΔU. The important thing is the (1/2)kz^2 term which produces simple harmonic motion with spring constant k. It produces the same oscillations as does (1/2)kx^2, but with the equilibrium point at z = 0 instead of at x = 0.

 

[PFuser1232]

x = z + c is not consistent with my diagram and my sign convention (which is that both x and z increase in the upwards direction). Looking at the first horizontal dashed line on my diagram, for example, if z = +2c, what do you have to do to it to get x = +c? (Your c is the same as my Δ).

But even after you fix that, U still ends up with a constant term along with the (1/2)kz^2. I must have made a mistake in my algebra yesterday (which I don't have here so I can't check it now). Sorry about that.

Nevertheless, that constant term is OK, because adding a constant to the potential energy everywhere doesn't change any physical results. What matters physically is the difference in potential energy, ΔU, between two different locations, and a constant term always cancels out when you calculate ΔU. The important thing is the (1/2)kz^2 term which produces simple harmonic motion with spring constant k. It produces the same oscillations as does (1/2)kx^2, but with the equilibrium point at z = 0 instead of at x = 0.

So what you mean to say is, the relationship U = 1/2 mω^2 z^2 is technically ΔU = 1/2 mω^2 z^2? And that I must specify my reference point in the equations for energy when it comes to SHM? Like, U = 1/2 kz^2 - 1/2 kc^2 is the total potential energy WITH RESPECT TO x=0 as the reference point, but I need an expression for U with respect to z=0 as the reference point?

 

[Nugatory]

So what you mean to say is, the relationship U = 1/2 mω^2 z^2 is technically ΔU = 1/2 mω^2 z^2? And that I must specify my reference point in the equations for energy when it comes to SHM? Like, U = 1/2 kz^2 - 1/2 kc^2 is the total potential energy WITH RESPECT TO x=0 as the reference point, but I need an expression for U with respect to z=0 as the reference point?

All potential energies are always specified relative to some reference point, always. We usually try to choose a reference point that makes the math simple (as when we decide to use the equilibrium position as the zero point), and sometimes the choice is so obvious that we're not even aware of it (when we choose the surface of the Earth as the zero point in some problems), but must always have a reference point.

 

[PFuser1232]

Yes I finally understood it, thanks everyone!
One last question, for a simple pendulum, when the small angle approximation is applied, how can we say that U = 1/2 kx^2 if the string is inextensible? Isn't U entirely U_grav in this case? And if it is, then we have a contradiction, because if the pendulum bob rises and descends periodically, this is in 2 dimensions not 1 dimension. And shm, according to my understanding, is a one dimensional phenomenon.

 

[jtbell]

Potential energy is always relative to some reference point where it is defined to be zero.

When only one kind of potential energy is involved, a particular reference point is often so customary that people don't bother to state it explicitly. With a spring, the customary reference point is the equilibrium point. With gravity near the Earth's surface, the customary reference point is the Earth's surface. With gravity in other situations, the customary reference point is often "at infinity".

Even if people don't state the reference point explicitly, you need to be aware of it, and if there are two or more forces involved, you need to make sure you're using the same reference point for both of them.

 

[olivermsun]

Yes I finally understood it, thanks everyone!
One last question, for a simple pendulum, when the small angle approximation is applied, how can we say that U = 1/2 kx^2 if the string is inextensible? Isn't U entirely U_grav in this case? And if it is, then we have a contradiction, because if the pendulum bob rises and descends periodically, this is in 2 dimensions not 1 dimension. And shm, according to my understanding, is a one dimensional phenomenon.

For a pendulum there is no spring. The "restoring force" is gravity itself, so the potential energy of the oscillator is just the gravitational potential energy mgz. The single variable is usually taken to be the angle of the pendulum.

 

[sophiecentaur]

For a mass on a spring (vertical set up) why is potential energy U defined as 1/2 kx^2? This is just the elastic potential energy. Shouldn't it be U = 1/2 kx^2 + mgh? Both the elastic AND potential energy? Also, for a simple pendulum at a very low amplitude, the potential energy is all gravitational, right? By the way, if U = 1/2 kx^2 + mgh, then the relationship ω = √(k/m) becomes invalid, since U = 1/2 mω^2 x^2 is always valid and equating this to both elastic and gravitational potential energy gives us a different expression for ω.

Potential Energy is always 'relative' to some position. That position is usually the equilibrium point, to make the calculations less complicated.
Why not solve it for yourself, including the mgh factor at the start? If you have done, then submit your workings here. Someone may help spot the error.