Energy, U, stored in springs as a function of y, the distance

  • #1

Homework Statement

The equilibrium length of each spring in the figure(attached, hopefully it shows up) is b, so when the mass m is at the center, neither spring exerts any force on it. When the mass is displaced to the side, the springs stretch; their spring constant is k.
(a) Find the energy, U, stored in the springs, as a function of y, the distance of the mass up or down from the center.
(b) Show that the period of small up-down oscillations is infinite.

Homework Equations

Potential Energy= mgy
or for springs=.5(ky^2)

The Attempt at a Solution

I really don't know how to go about it, but I've tried using a different equation that our professor showed us that goes (1/2)kx^2 + bx^4. I'm not understanding where the initial and finals are located. Also, the problem says it is stretched to the side so that would indicate a horizontal plane in which the gravitational energy is constant and can be left out, right? It is a symbolic problem so this is all that is given and the answer should be in Joules I believe or Nm-1. I have attached the figure for the problem. Any sort of help would be greatly appreciated. Thanks in advanced! :shy:


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    ch2 p36.jpg
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Answers and Replies

  • #2
The spring oscillating up and down forms a right triangle. The displacement from equilibrium, is y and is the vertical leg of the right triangle the equilibrium length of the spring is b and is the horizontal leg of the right triangle. Then the length of the spring when it is stretched up or down is the hypotenuse, given by,

(y^2 + b^2)^(1/2)

remember that the half power just means square root.
So if we consider how far the spring is stretched it is always given by, its stretched length minus its equilibrium length.


change in length = (y^2 + b2)^(1/2) - b

This then gives us a formula for the energy in 1 spring as,

U = (1/2)*k*[(y^2 + b^2)^(1/2) - b]

then the total energy in the two springs is,

U = k*[(y^2 + b^2)^(1/2) - b]

Thats the answer to part one.

Second we must show for small oscilations that the period is infinite.

now lets sum the forces on the mass at the peak of its motion.

let A be the angle between the hypotenuse and the leg b. then the displacement, y, is given by,

y = b tan(A)


Cos[A] = b / hypotenuse

hypotenuse = b / cos[A]

so we can write the spring force equation as

k*(change in spring length) = m*a(y-direction)

since the spring pulls back along the direction of the hypotenuse we must divide its force into components to find the total force in the y-dirction, and get the differential equation for the y motion. Since the x direction forces are equal and opposite, they will cancel. so for the vertical force we have,

-2k*(b/cos[A] - b)sin[A] = m *y''

Note the two comes from the fact that we have 2 springs, and hence two vertical components.
here we can write y = b*tan[A]

and for very small A, 1>>>>A

tan[A] = A

so y'' = b A''

and the differential equation becomes,

-2k*(b/cos[A] - b)sin[A] = m *b*A''

here the term -2k*(b/cos[A] - b) is like the spring constant for the group of springs, so the frequency of oscillations is,

w = [-2k*(b/cos[A] - b)/m]^(1/2)

but we also note that


sin[A] = A
Cos[A] = 1

so the equation reduces to,

-2k*(b-b)A = m*b*A''

If youll notice the constant in front of A is zero. so

w = [-2k*(b - b)/m]^(1/2)=[0/m]^(1/2)=0

w = 2*PI*frequency = 2* PI/Period

0 = 2* Pi/period

1 / period = 0

so the period must be infinite.

So this ends the problem.
I have included a small section in slightly more depth. If you are fine with just the basic solution you dont have to read on.

If you are interested,
as a side note, this is only true in an approximation.

we may write

cos[A] = 1 - (A^2)/2! + (A^4)/4! - ...

this is done using the taylor series you may remember from

Then if we take the second order terms only since when A is close to zero, A^4, A^6, A^8.... are all really small.

we can say,

cos[A] = 1 - (A^2)/2!


-2k*(b/(1 - (A^2)/2!) - b)sin[A] = m *b*A''

then if we recall, 1/(1 - (A^2)/2!)= 1+[A^2]/2

and 2! = 2*1 = 2

and sin[A] = A for small angle

we can write,

-2k*(b(1 + (A^2)/2!) - b)A = m *b*A''

which reduces to,

-2k*b(A^3)/2!) = m *b*A''

this can be integrated if you note,

A'' = dA'/dt = [dA'/dA] * [dA/dt]

You can just think of this as fractions and the dA's can cancel
so this implies,

A'' = [dA'/dA]*A'

and then

-2k*b(A^2)/2!)A = m *b*[dA'/dA]*A'

-2k*b(A^3)/2!)dA = m *b*A'*dA'

we can now integrate both sides since we only have A terms on the left side, and A' terms on the right.

integrating both sides
we get,

-2k*b(A^4)/6)+C = (1/2)*m *b*(A' )2

where c is an integration constant. solving for A' i get,

A' = [-2k(A^4)/(3m) + 2C]^(1/2)

If we say then that the speed was zero intially then A' must be zero when we plug in out initial position A0 (read A-zero) in for A on the right. We can then see that C must be

C = k*(A0^4)/(3m)

and now we can write,

A'= [-2k(A^4)/(3m) + 2k*(A0^4)/(3m)]^(1/2)
A'= [-2k(A^4-A0^4)/(3m)]^(1/2)

this is where I would stop as further approximations would lead bad results and its hard to seperate this for another integration with a square root on the right side.

I don't know how far into the problem you wanted to go. I thought id include this last section incase you were really interested in physics. I would encourage you to go into the field if you liked this stuff. If you didn't like it, or it was over your head, dont worry. I've been studying this stuff for 6 years, and if you decided to focus in this field, they would teach it to you slowly and build you up to this level. I hope this was helpful.

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