Magnet on strap (also under Systems > Rotating Strap)

Suppose a 50 gram magnet is 20 cm from the axis of rotation, with the strap moving at 50 degrees / second.

The magnet travels in a circular path of radius 20 cm.  The circumference of that path is 2 pi r = 2 pi * 20 cm = 40 pi cm, about 125 cm.

50 degrees is 50/360 of a complete rotation (360 deg for one rotation).

So during a 50 degree rotation the magnet moves 50/360 * 40 pi cm = 18 cm, roughly.

The magnet is therefore moving at about 18 cm/second.

University Physics Notes (also under Concepts > Pendulum and Systems > Pendulum)

A pendulum or a mass on a spring is influenced by a net force which is proportional to distance from equilibrium, acting in the direction of equilibrium.  A force of this nature is described by

The buoyancy-stabiliized balance is characterized by restoring torque of similar nature.  More about this later, but the mathematics for the two systems are nearly identical.

Now starting with F_net = - k x we use Newton's Seconds law to rewrite this as

Since a = dv/dt and v = dx/dt, we conclude that a = x '', where ' indicates derivative with respect to t.  Our equation becomes

To solve this equation we need to find a function x(t) which satisfies this equation. 

The function must have the characteristic that its second derivative is a constant multiple of the itself.  We know three functions with this characteristic:

An exponential function of the form x(t) = A e^( c t ), with c a constant, has second derivative c^2 A e^ (c t).  Thus x '' = c^2 * x.

A sine function of the form x(t) = A sin(c t) has second derivative - c^2 A sin(c t).  Thus x '' = - c^2 x.

A cosine function of the form x(t) = A cos(c t) has second derivative - c^2 A sin(c t).  Thus x '' = - c^2 x.

If we use the exponential function in our equation x '' = -k / m * x, we get the equation

c^2 A e^(c t) = -k / m * A e^(c t), which simplifies to

c^2 = - k / m, with solution

c = +- sqrt(k / m) * i, where in this case i = sqrt(-1).

The imaginary solution would seem to be an obstacle, but in fact it isn't.  I've sketched a few details below, but they aren't necessary for your work right now (worth knowing for your upcoming diff eq course).  These details can be used as an overall reference later in this course:

Euler's formula

is obtained by doing a Taylor expansion of the exponential (e^x = 1 + x + x^2 / 2! + x^3 / 3! + ... ), and substituting (i theta) for x.  When you simplify the resulting expression and separate the real and imaginary parts the Taylor expansions for cosine(theta) pop right out at you.

Substituting c = sqrt(k/m) * i into x = A e^(c t) we get x = A (cos( sqrt(k/m) t) + i sin(sqrt(k/m) t).

Either the real or the imaginary part of the solution will work for the equation.  So will any linear combination of the two. 

Conclusion: 

is the general solution to the equation.

This solution can be put into the form x(t) = A sin( omega t + phi), where A and phi can be any constants.  A is interpreted as amplitude, phi as the starting phase of the motion.

If we use the cosine function in the equation x '' = -k/m * x we get the equation

-c^2 A cos( c t) = -k/m A cos(ct)

so that -c^2 = -k/m and c = +-sqrt( k/ m).

Our solution becomes x(t) = A cos( sqrt(k/m) * t)

This solution can be interpreted in terms of the circular model of the sine and cosine functions.  This description goes beyond what you are required to know right now, but can serve as a reference for present and future work with simple harmonic oscillators:

x = A cos(theta) is the x component of the point on a reference circle of radius A, where the radial vector from the center of the circle to the point makes angle theta with the positive x axis.

x(t) = A cos( omega * t) is therefore the x component of the point on a reference circle of radius A, where the radial vector from the center of the circle to the point makes angle theta = omega * t with the positive x axis.

If t is interpreted as time, then theta = omega * t gives the angle of the radial vector whose angular velocity is omega.  The projection of this vector onto the x axis models the position of the oscillating object as a function of time.

In our solution function x(t), which involves cos(sqrt(k/m) * t), omega is just sqrt(k/m).  That is:

omega = sqrt( k / m ).

In a nutshell:

x(t) is the x projection of the vector r(t) of constant magnitude A whose angular position is given by the function theta = omega * t.

Consequences:

The reference point has constant speed omega * A, and it at any instant moving in the direction tangent to the circle. 

The direction tangent to the circle is perpendicular to that of the position vector r(t). So the velocity of the reference point is described by the vector v(t) whose magnitude is omega * A and whose direction is perpendicular to the r(t) vector.

The reference point has an acceleration toward the center of the circle, whose magnitude is a = v^2 / r, where v and r are the magnitudes of the velocity and position vectors v(t) and r(t).  Those magnitudes are omega * A and A, so the magnitude of the acceleration can also be describe by a = omega^2 * A. 

Thus we can sketch the reference circle, with r being the radial vector from center to the reference point, v perpendicular to r representing the velocity of the reference point and a perpendicular to v directed toward the center of the circle.

The projection of the r vector in the x direction is the corresponding position of the oscillating point, the projection of the v vector in the x direction is the corresponding velocity of that point, and the projection of the vector a in the x direction is the corresponding acceleration of that point.

These projection vectors will be identical to v_x(t) = x ' (t) and a_x(t) = x '' (t).  Remember that x(t) = A cos(omega * t).

This gets even better.  We can project the r, v and a vectors in any fixed direction we choose.  Think of projecting these vectors onto a directed line through the center--any directed line through the center.  (If that line happens to be the y axis, then we will have y(t) = A sin(omega t), and the velocity and acceleration functions will be the projections v_y(t) = y ' (t) and a_y(t) = y '' (t).  If the line happens to be in some other direction, then the position function will be a linear combination of the sine and cosine functions, as will the velocity and acceleration functions).