# 15.4: Comparing Simple Harmonic Motion and Circular Motion

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##### Learning Objectives
• Describe how the sine and cosine functions relate to the concepts of circular motion
• Describe the connection between simple harmonic motion and circular motion

An easy way to model Simple Harmonic Motion (SHM) is by considering uniform circular motion. Figure $$\PageIndex{1}$$ shows one way of using this method. A peg (a cylinder of wood) is attached to a vertical disk, rotating with a constant angular frequency.

Figure $$\PageIndex{2}$$ shows a side view of the disk and peg. If a lamp is placed above the disk and peg, the peg produces a shadow. Let the disk have a radius of r = A and define the position of the shadow that coincides with the center line of the disk to be x = 0.00 m. As the disk rotates at a constant rate, the shadow oscillates between x = + A and x = −A. Now imagine a block on a spring beneath the floor as shown in Figure $$\PageIndex{2}$$.

If the disk turns at the proper angular frequency, the shadow follows along with the block. The position of the shadow can be modeled with the equation

$x(t) = A \cos (\omega t) \ldotp \label{15.14}$

Recall that the block attached to the spring does not move at a constant velocity. How often does the wheel have to turn to have the peg’s shadow always on the block? The disk must turn at a constant angular frequency equal to 2$$\pi$$ times the frequency of oscillation ($$\omega$$ = 2$$\pi$$f).

Figure $$\PageIndex{3}$$ shows the basic relationship between uniform circular motion and SHM. The peg lies at the tip of the radius, a distance A from the center of the disk. The x-axis is defined by a line drawn parallel to the ground, cutting the disk in half. The y-axis (not shown) is defined by a line perpendicular to the ground, cutting the disk into a left half and a right half. The center of the disk is the point (x = 0, y = 0). The projection of the position of the peg onto the fixed x-axis gives the position of the shadow, which undergoes SHM analogous to the system of the block and spring. At the time shown in the figure, the projection has position x and moves to the left with velocity $$v$$. The tangential velocity of the peg around the circle equals $$\bar{v}_{max}$$ of the block on the spring. The x-component of the velocity is equal to the velocity of the block on the spring.

We can use Figure $$\PageIndex{3}$$ to analyze the velocity of the shadow as the disk rotates. The peg moves in a circle with a speed of vmax = A$$\omega$$. The shadow moves with a velocity equal to the component of the peg’s velocity that is parallel to the surface where the shadow is being produced:

$v = -v_{max} \sin (\omega t) \ldotp \label{15.15}$

It follows that the acceleration is

$a = -a_{max} \cos (\omega t) \ldotp \label{15.16}$

##### Exercise 15.3

Identify an object that undergoes uniform circular motion. Describe how you could trace the SHM of this object.

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