One Dimensional F(x)

Basics

Newton’s Second Law:   F(x)  =  m d2x/dt2  =  m dv/dt

Unlike Newton’s 2nd Law with F(t) and F(v), this equation doesn’t easily separate. Trick: If we multiply both sides by "v" we get: F(x)v  =  mv dv/dt ,

and with v = dx/dt this becomes:  F(x) dx/dt  =  m v dv/dt .

Now multiply through by dt to get:  F(x) dx  =  m v dv .  We have now separated the equation with only x’s on one side and v’s on the other!

We can now integrate: xoò x F(x) dx  =  ½mv2 - ½mvo2 .

In two or more dimensions, we have to worry about the path, but in one dimension there is only one path.

The integral on the left is the work done, and we can call it a potential energy, V(x), if that work results in stored energy (as opposed to a dissipative energy like heat from friction)

V(x) - V(xo)  =  - xoò x F(x) dx . Note that an integration has to be performed.

Newton's Second Law with F(x) can now be written as:

½mv2 + V(x)  =  ½mvo2 + V(xo)  =  constant  =  E (where we call E the energy).

We can solve for the velocity, v, but this gives us v as a function of x, not t:  v(x)  =  [2{E-V(x)}/m]1/2  =  dx/dt .

We now separate the equation: all the t’s on the left and all x’s on the right: dt  =  dx / [2{E-V(x)}/m]1/2 .

Integrating gives:  t - to  =  [m/2]1/2 xoò x [1 / {E-V(x)}]1/2 dx .

Note that we have a second integral to do.  Also note that both xo and vo are involved in the integrations.  Having two integrations to do and having two initial conditions (xo and vo) are common to our solutions for F(t), F(v), and now F(x).

Assuming we can do both integrals, we then get t(x) rather than x(t). To get the final x(t), we need to be able to invert t(x) to get x(t). This may or may not be easy to do.

Homework Problem: Problem #7:  Find x(t) and v(t) for a particle of mass, m, projected up from the earth’s surface with the escape speed for the earth.  Neglect air resistance.

Potential Energy and F(x)

If we choose some standard position, we can define a potential energy function:  V(x) - V(xs) = - xsò x F(x) dx

By adding and then subtracting V(xs) to our energy equation, we get:

V(x) - V(xs) + V(xs) - V(xo)  =  ½mv2 - ½mvo2  =  [V(x) - V(xs)] - [V(xo) - V(xs)] .

From the above it is easy to see that the xs position really doesn't matter at all. This is why PE = mgh can be used as long as we pick some "nice" position to be h=0. It can be the ground, it can be the floor, or the desk top. It really doesn't matter - as long as we consistently measure h from the same position throughout the problem.

Now by having a definite V(x), we can plot V(x) versus x. We see from the equation:

½mv2 + V(x)  =  ½mvo2 + V(xo)  =  constant = E ,   or   ½mv2 + V(x)  =  E .

Note that since the kinetic energy, ½mv2 , is always positive, the only allowed positions are those positions that have V(x) less than (under) the value of E. Any point on the V(x) plot that crosses the E=constant line will have v=0 at that position - hence this will be a turning point of the motion. We will draw some V(x) curves in class and see what they tell us about the motion.

Example: Harmonic Oscillator.

This is a nice example and an important example as we will shortly see.

F(x) = -kx (spring). From experience we know that a mass on a spring will oscillate, so we might suspect that the solution for x(t) will involve a sine function.

If we started from Newton's Second Law directly, we would have:   -kx(t)  =  m d2x(t)/dt2

By our knowledge of functions, we could guess that x(t)  =  A sin(wt + qo) since the second derivative of a sine function gives itself back again with a negative sign. And since cosine also behaves this way, we might choose it. But including an initial phase angle, q o , we can take care of both the sine and cosine possibilities.  To see this, recall the trig identity:  sin(a+b)  =  sin(a)cos(b) + cos(a)sin(b), so that A*sin(wt + qo)  =  A*cos(qo)*sin(wt) + A*sin(qo)*cos(wt)  =  B*sin(wt) + C*cos(wt)  where  B = A*cos(qo) and C = A*sin(qo) .  In both cases, we have two constants, (A and qo) or (B and C), to take care of the two initial conditions, usually (xo and vo).

But let's see how this works out analytically using the above energy procedure:

Step 1:  Define a potential energy:

V(x) - V(xs)  =  - xsò x (-kx) dx  =  ½kx2 - ½kxs2 .

If we let xs = 0, then V(xs) = 0 and V(x) = ½kx2 .

Step 2:  Solve for v(x) using Conservation of Energy:

½mv2 + ½kx2  =  E,   so  dx/dt  =  v  =  [2{E-V(x)}/m]1/2 .

Step 3:  Separate x and t, and integrate to try to find t(x):

to=0ò t dt  =  xoò x dx / [2{E-V(x)}/m]1/2  ,  or

t(x)  =  (m/2)1/2 xoò x dx / [E - ½kx2]1/2

Now we take the E out of the [ ]1/2 to get:   t(x) = (m/2E)1/2 xoò x dx / [1 - ½kx2/E]1/2

Now we set y2 = (k/2E)x2 which makes 2y dy = 2(k/2E)x dx , or

dx  =  2y dy / [2(k/2E)x]  =  2 (k/2E)1/2x dy / [(k/E)x]  =  (2E/k)1/2 dy so that

t(x)  =  (m/2E)1/2 xoò x dx / [1- ½ kx2/E]1/2

t(x)  =  (m/2E)1/2 (2E/k)1/2 yoò y dy / [1-y2]1/2  =  (m/k)1/2 yoò y dy / [1-y2]1/2

where yo = [k/2E]1/2 xo  and  y = [k/2E]1/2 x .

Now we let y = sin(q ) so that dy = cos(q ) dq we have:

t(x)  =  (m/k)1/2 qoò q cos(q ) dq / [1-sin2(q )]1/2  =  (m/k)1/2 qoò q dq   =  (m/k)1/2 (q -q o)

where q  =  sin-1(y)  =  sin-1([k/2E]1/2 x) . Thus we have:

t(x)  =  (m/k)1/2 sin-1([k/2E]1/2 x) - q o .

Step 4:  Invert t(x)  to try to get x(t):

x(t)  =  A sin(w t + q o)  where w = (k/m)1/2  and  A = [E/2k]1/2 .

This is the result we got above by our knowledge of functions.

Analysis:

As k increases, the frequency of the oscillation increases.

As m increases, the frequency of the oscillation decreases.

The amplitude depends on the energy (but as the square root of the energy - not linearly).