# A mass m sits at x = 1, which is the neutral position of two springs, the first of which connects the mass to the point x = 2 and has spring constant k1, and the second of which connects the mass...

A mass m sits at x = 1, which is the neutral position of two springs, the first of which connects the mass to the point x = 2 and has spring constant k1, and the second of which connects the mass to the origin and has spring constant k2. At t = 0 the mass is pulled to x = 1/2 and released. There is no gravity and no friction and the mass moves back and forth only on the x-axis. Where is the mass at time t? Hint: use the Euler-Lagrange equation applied to the function (kinetic energy minus potential energy). Kinetic energy is the usual (1/2)*m*v^2 and potential energy is elastic potential energy only. The position function x(t) will be the solution to the differential equation. (You can solve the diffEQ via Laplace if you want).

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Note: This might be a different way to solve this, you can decide if it is valid within your question restraints.

The following assumptions must be true:

The spring must follow Hooke's Law

The spring is of negligible mass

The system is completely friction free (gravity could be involved, but the force will be balanced by the surface that this mass might be resting on - either way it doesn't matter)

If we release a mass on such a spring, the acceleration will be found using Newton's second law:

`ma=-k_sx`

Since this is a system of two springs that are at equilibrium when x=1, we then construct a force diagram to see what the net force is:

Spring 1 stretched from zero. The amount of stretch then is based on x directly.

`F_1=-k_1x`

Spring 2 stretched from 2. The amount of stretch then is based on (x-2).

`F_2=-k_2(x-2)`

Second law statest that` F_1+F_2 = ma` , which gives us:

`-k_1x+(-k_2(x-2))= ma`

We know that at x=1, the system is in equilibrium meaning the net force is zero.

`-k_1-k_2(-1) = 0 rArr k_1=k_2=k`

Plugging back into our acceleration equation we get:

`-kx-k(x-2) = ma`

`-2kx+2k = ma`

`or -2k(x-1)=ma`

`text(let) k_s = 2k`

so `-k_s(x-1)=ma`

acceleration is the second derivative of x (`ddotx`) and x is a function of time x(t).

so `a = -k_s/m(x-1) rArr ddotx = -k_s/m(x-1) rArr ddotx+k_s/mx-k_s/m=0`

This differential equation can be solved using a bit of experimental knowledge. Since this is an oscillating system, what functions oscillate? Sinusoidal.

the generic function for a sinusoidal is

`x(t)=Acos(omegat+phi)+x_0`

where `omega=(2pi)/T` , such that when t = T (1 period), the function will have completed one cycle and, in this case, `x_0 = 1` . Now, let's start doing the problem:

Since we are releasing this from rest: when t = 0, v = 0

`v = dx/dt or v= dotx` .

`dx(t)=d(Acos(omegat+phi)+1)rArrdotx=-Aomegasin(omegat+phi)`

when `v(0)=0` so, `0=-Aomegasin(phi)`

Since `Aomega` cannot be zero, then `sin(phi)=0` hence, `phi=0` .

That reduces our problem a bit

`x = Acos(omegat)+1rArr x-1 = Acos(omegat)`

`dotx=-Aomegasin(omegat)`

Thus, `ddotx=a=dv/dt=d(-Aomegasin(omegat))rArrddotx=-Aomega^2cos(omegat)`

Since `(x-x_0)= Acos(omegat)` , then `ddotx = -omega^2(x-1)`

Substituting this into our original equation we get:

`ddotx+k_s/mx+k_s/m=0rArr-omega^2(x-1)+k_s/m(x-1)=0`

Thus `omega=k_s/m`

Now we are left with finding A.

`x(t)=Acos(k_s/mt)+1`

At t=0, x=0.5

`x(0)=0.5-Acos(0)+1`

`:.A=0.5`

So our final equation, substituting back `k_s=2k_1` :

**Answer:** `x(t)=0.5cos((2k_1)/mt)+1`