Solving for λ, a, and Vmax: An Analysis of Energy Balance and SHM

[Naraliya]

Homework Statement

A particle P of mass m, is attached to one end of a light elastic spring of natural length a and modulus of elasticity kmg.

The other end of the spring is attached to a fixed point O on a ceiling.

The point A is vertically below O such that OA = 3a.

The point B is vertically below O such that OB = 1/2a.

The particle is held at rest at A and the released, coming to instantaneous rest at the point B.

a) Show that k = 4/3

b) Find, in terms of g, the acceleration of P immediately after being released from rest at A.

c) Find, in terms of g and a, the maximum speed attained by P as it moves from A to B.

Relevant Equations

F = ma

T = (lamda * x)/natural length) tension/thrust in a spring/string

E.E =( lamda * (x)^2)/(2 x natural length) elastic potential energy

a = -omega * x S.H.M

m * g * h = G.P.E gravitational potential energy

a) λ = 4/3 by considering the energy balance as P moves from A (2a) to B (1/2a). The E.E. at A changes into a gain in gravitational potential energy + build up of E.E. at B since the spring compresses.

b) a = 5/3g by considering that the mass P is in dynamic equilibrium immediately after release. So, F = ma = T - mg = ma. Tension in the string is due to the extension to A.

c) This is where I'm stumped. I've tried many different ways and get different answers. I KNOW that the maximum speed occurs at the equilibrium position where the acceleration = 0. If the natural length of the spring is 1a, then the "extension" of the spring works out to be 3/4a, which makes the equilibrium position at 7/4a (1+3/4).

Doing another energy balance from A to this equilibrium and ignoring the fact the eventually the mass will compress the spring, I get a Vmax of root(29/6 * g * a).

But then I start thinking deeper and considering S.H.M. There must be SHM around the equilibrium position 7/4a, and when the mass goes up past it, eventually the spring gets compressed and you get a build up of thrust until the mass reaches 0 speed at B (1/2a). Every permutation of me using SHM equations ma = -ω^2x and then using Vmax = ωx, gives me a different value of Vmax.

I don't know what is the right answer and whether I'm overthinking it (should I only care about the motion from A to the equilibrium position) or whether I'm leaving something out (do I factor in that at B = 1/2a, the spring is compressed and providing a thrust force?).

Please help, I've spent days on this now.

 

[kuruman]

You can treat a vertical spring exactly the same as a horizontal spring. The mechanical energy is still $ME=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ and $x$ is still the displacement from the equilibrium position. In the horizontal case the equilibrium position is at the unstretched position of the spring; in the vertical case is where the spring (upward) force balances the (downward) weight. I hope this helps.

 

[Naraliya]

$\frac {4/3 m g (2a)^2} {2a} = 5/4mga + 1/2mv^2$

Elastic energy in spring at A = gain in potential to equilibrium point (3a to 7/4a) + kinetic energy built up in the mass during that motion.

I get an answer of $\sqrt{17/6ga}$ as the maximum speed. I don't think that's correct.

From thinking about it again, there is SHM around the equilibrium point. Since the object was at rest and released from A(3a), then comes to rest at B(1/2a), equilibrium point is 5/4a. $\omega$ I found to be $\sqrt{ \frac {4/3g}{a} }$. So from that method I get a speed of $\sqrt{25/12ga}$.

You can see my frustration now as every answer seems different and I'm not sure which one is correct.

 

[kuruman]

This equation $\frac {4/3 m g (2a)^2} {2a} = 5/4mga + 1/2mv^2$ is dimensionally incorrect and does not make sense. If you want to use energy conservation, you need to know the amplitude of oscillations $A$ and the spring constant $k$. Then write expressions for the kinetic and potential energy.

 

[Naraliya]

Why doesn't it make sense? I wrote elastic energy = gravitational potential energy + kinetic energy. I can't see how its not dimensionally correct.

And I know the amplitude of the oscillations. I worked out the equilibrium point and confirmed that the object should perform SHM between point A and B about that point. The amplitude works out to be 5/4a. When released from point A, even if I take A as zero gravitational potential, it still has elastic potential due to the stretched spring. So that elastic potential raises the mass to a height (lets say to the equilibrium point) and also gives it kinetic energy.

I really can't see why that thinking is wrong.

 

[kuruman]

Sorry, the equation is actually dimensionally correct. I agree that the amplitude is $A=\frac{5}{4}a$. What is the spring constant $k$?

 

[Chestermiller]

If I write a force balance on the mass (after it has been released), I get:
$$mg-kmg\frac{(x-a)}{a}=m\frac{d^2x}{dt^2}$$where x is the distance measured downward from the ceiling. If I multiply this by dx/dt, I obtain:
$$mg\frac{dx}{dt}-kmg\frac{(x-a)}{a}\frac{dx}{dt}=m\frac{dx}{dt}\frac{d^2x}{dt^2}$$If I next integrate this from time zero to time t, I obtain: $$mg(x-3a)-kmg\frac{(x^2-2ax-3a^2)}{2a}=m\frac{v^2}{2}$$What is k equal to if v = 0 when x = a/2?

 

[Naraliya]

k = -10/3 when I plug in x = a/2 and v =0...

My brain is exploding.

 

[Naraliya]

Sorry, the equation is actually dimensionally correct. I agree that the amplitude is $A=\frac{5}{4}a$. What is the spring constant $k$?

This is UK A-level Maths (mechanics). We don't deal with spring constants right now as our form of hooks law is actually $T = \frac{\lambda x}{l_0}$ where $\lambda$ is the modulus of elasticity.

But if that really is the amplitude of the motion between points A and B, then maximum speed on the particles way to B would be $V_{max} = \omega A$ right? $\omega$ is found using the expression $ma = -\frac{4/3mg(e+x)}{a} + mg$

 

[Chestermiller]

k = -10/3 when I plug in x = a/2 and v =0...

My brain is exploding.

Try again. It comes out to 4/3

 

[haruspex]

Try again. It comes out to 4/(3a)

k does? k should be dimensionless.

 

[Chestermiller]

k does? k should be dimensionless.

I had interpreted kmg as the spring constant which has units of force per unit length. But, I see now what it is. I've gone back and corrected this in my previous posts.

 

[haruspex]

But if that really is the amplitude of the motion between points A and B, then maximum speed on the particles way to B would be $V_{max} = \omega A$ right? $\omega$ is found using the expression $ma = -\frac{4/3mg(e+x)}{a} + mg$

This is getting confusing with A being both a point in space and an amplitude, a being both a length and an acceleration. e, I assume, is equal to a (length).
Which way are you defining as positive in that equation?

 

[Naraliya]

I interpreted kmg as the spring constant which has units of force per unit length.

kmg is the modulus of elasticity, where k = 4/3 and is dimensionless.

 

[Chestermiller]

Sorry, the equation is actually dimensionally correct. I agree that the amplitude is $A=\frac{5}{4}a$. What is the spring constant $k$?

I get an equilibrium point of x = (3a+a/2)/2=1.75a and an amplitude of A=(3a-1.75a)=(1.75a-0.5a)=1.25a

 

[Chestermiller]

kmg is the modulus of elasticity, where k = 4/3 and is dimensionless.

OK. I see what this represents. I've gone back and modified my previous posts to take this into consideration. Please consider my previous posts again with these changes implemented.

 

[Naraliya]

I get an equilibrium point of x = (3a+a/2)/2=1.75a and an amplitude of A=(3a-1.75a)=(1.75a-0.5a)=1.25a

So I'm not crazy then! And the max speed is simply V = wA?

 

[Chestermiller]

So I'm not crazy then! And the max speed is simply V = wA?

This is certainly going to be true. The maximum velocity can also be determined by setting x equal to the equilibrium distance x = 1.75a in my third equation in post #7.

 

[Naraliya]

This is certainly going to be true. The maximum velocity can also be determined by setting x equal to the equilibrium distance x = 1.75a in my third equation in post #7.

It works! And it agrees with the answer I got when I used v = Aw! I can rest now.

But first, time for some plenary learning. Your first force balance equation, what direction are you taking it from? From the ceiling or from point A which is 3a below the ceiling? Because I don't understand why the force balance has mg as the resultant force.

 

[Chestermiller]

It works! And it agrees with the answer I got when I used v = Aw! I can rest now.

But first, time for some plenary learning. Your first force balance equation, what direction are you taking it from? From the ceiling or from point A which is 3a below the ceiling? Because I don't understand why the force balance has mg as the resultant force.

The force balance is in the downward direction, and x is the position of the particle measured downward from the ceiling. The downward force of gravity on the mass is mg and the upward force of the spring on the mass is $kmg\frac{(x-a)}{a}$. So the net downward force on the mass is $mg-kmg\frac{(x-a)}{a}$, which is equal to $m\frac{d^2x}{dt^2}$. Of course, the two force terms can be combined to give $-kmg\frac{(x-a-a/k)}{a}$ so that it is expressed relative to the equilibrium position.

 

[Naraliya]

The force balance is in the downward direction, and x is the position of the particle measured downward from the ceiling. The downward force of gravity on the mass is mg and the upward force of the spring on the mass is $kmg\frac{(x-a)}{a}$. So the net downward force on the mass is $mg-kmg\frac{(x-a)}{a}$, which is equal to $m\frac{d^2x}{dt^2}$. Of course, the two force terms can be combined to give $-kmg\frac{(x-a-a/k)}{a}$ so that it is expressed relative to the equilibrium position.

Thank you, that makes sense.

For your 2nd equation, what methods of integration did you use to get the 3rd? I can just about argue the first term because at t = 0, x is variable x to be incremented along until you get to t = t at which x becomes 3a. Not sure how you integrated the rest of the terms, especially the 2nd derivative on the right hand side.

 

[Chestermiller]

Thank you, that makes sense.

For your 2nd equation, what methods of integration did you use to get the 3rd? I can just about argue the first term because at t = 0, x is variable x to be incremented along until you get to t = t at which x becomes 3a. Not sure how you integrated the rest of the terms, especially the 2nd derivative on the right hand side.

At time zero, x =3a, and at time t, x is x. So the limits of integration on x are from 3a to x.
On the right side you have vdv