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# Simple Harmonic Motion Study Guide (page 2)

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Updated on Sep 27, 2011

## Period, Frequency, and Speed of Oscillatory Motion

In order to have a visual representation of the quantities involved in the simple harmonic motion, let's look at Figure 16.2. Imagine the object oscillating around the equilibrium position and finally touching and making an impression on a screen that is moving horizontally. What is the final trajectory of the object on the screen? If we analyze Figure 16.2, we will see that the elongation or displacement varies with time in a repetitive way. We say we have an oscillation. One complete oscillation takes us from one point on the graph to another completely identical point further along. One example is shown in the figure: Points A and B are 1 oscillation apart. This is called a cycle.

This is similar to the uniform rotational motion and the related quantities we defined. Imagine a platform rotating uniformly and a side light shining on an object on the platform. The shadow of the object is projected on a moving screen in order to obtain a trajectory for this motion. See Figure 16.3.

The image obtained on the screen is very similar to the one we had shown previously for the moving spring, and it describes the oscillation of the object.

In the representation in Figure 16.4, the maximum displacement is achieved when the object passes through points A and –A away from the origin, called also amplitude. Note that at t = 0 seconds, displacement is zero.

The shape of the graph is also similar to the angle dependence of the simple trigonometric functions: sin θ will start at 0 for zero angle, and will reach + 1 when the angle is 90° and –1 when the angle is 270°. Therefore, the displacement can be written in terms of the amplitude A and as a function of the sine if we can determine the angle characterizing the oscillatory motion.

x ~ A (you would read: Displacement is proportional to amplitude)

and

x ~ sin (angle)

For this proportionality, the last representation of the uniform circular motion is very useful. If we consider starting at time t = 0 seconds and consider the angular displacement of the object on the platform is 0°, then after a time equal with one period of rotation, T, the angular displacement is 2π. Hence, the angular speed is:

as defined previously for uniform circular motion. The angular displacement after a time t will be:

θ = ω · t

And the desired trigonometric function to relate the uniform circular motion to the projection on the screen is:

sin (ω · t)

Hence the time dependence of the displacement can be written as:

x (t) = A · sin (ω · t)

One can see that at the initial time when t = 0 seconds, angular displacement is zero and x is zero. At 90°, angle displacement is + A, and at 270°, displacement is –A.

#### Example 1

Determine the amplitude A of an oscillatory motion if the displacement is 20.0 cm between times t0 = 0 seconds and time t = T/3, where T is the period of oscillations.

#### Solution 1

Convert units to SI and set the equations to determine the unknown.

Δx = xx0 = 20 cm = 0.20 m

t0 = 0

t = T/3

A = ?

0.2 m = A · sin (ω · t) – A · sin (ω · t0)

0.2 m = A· [sin (ω · t) – sin (ω · t0)]

0.2 m = A · [sin (ω · T/3) – sin (ω · 0)]

0.2 m = A · [sin (ω · T/3) – 0]

0.2 m = A ·sin(ω · T/3)

We can replace the angular speed with its previous definition

and:

0.2 m = A · sin [(2 · π/T) · T/3]

0.2 m = A · sin (2 · π/3 )

0.2 m = A · 0.87

A = 0.2 m/0.87 = 0.23 m

Hence, the amplitude of oscillation is:

A = 0.23 m

The inverse of the period is frequency, and it represents the number of oscillations completed in a unit of time (1 second).

f =

Therefore, the angular speed, also called angular frequency, can be expressed as:

ω = = 2 · π · f

The unit for the angular speed is radian/second (rad/s), whereas frequency is measured in hertz (Hz = 1/s), and period is in seconds.

#### Example 2

For the previous example, find the period and frequency if the angular speed is 2 degrees per second. Find the angular speed in SI.

#### Solution 2

First, set your units in SI and then solve for the unknowns according to the definitions.

ω · = 2° = 2° · π/180° = π/90 rad/s

ω · = 3.5 · 10–2 rad/s

T = ?

f = ?

T = 2 π/ω

T = 2· π /3 · 5 ·10 2 rad/s

T = 180 s

Then, frequency is nothing other than the inverse of the period:

f = vT = 1/180s

f = 5.5 · 10 –3 Hz

Tangential speed of the uniformly rotating object can be used to determine the speed for the simple harmonic oscillator. In the uniform circular motion, the tangential speed and the angular speed are related as:

vT = ω · r

where r is the radius of the circular motion.

The diagram in Figure 16.5 shows the speed of the oscillator to be the vertical component of the tangential speed (the projection on the screen). For the far left point, that is exactly the tangential speed (the projection of vT is the speed itself). While moving away from the equilibrium position, the speed decreases (projection is smaller), and when the object is at maximum displacement, the speed becomes zero only to increase when the object returns to the equilibrium position (although it now has a negative sign because this time, it is approaching the origin contrary to before when it was moving away).

The vertical component of the speed is seen to be:

v = vT · cos(θ)

But because the object experiences a uniform circular motion, the angular displacement is given by the angu- 1ar speed:

v(t) = vT · cos (ω · t)

Or:

v(t) = ω · r · cos (θ) = ω · r · cos (ω · t)

If you also compare the circular motion to the shadow the object leaves on the screen, one can determine the radius to be the amplitude of the oscillatory motion A.

r = A

And then the final formula for the speed is:

v(t) = ω · A · cos (θ) = ω· A · cos (ω · t)

Even more interesting is to see that we start with a uniform rotational motion (speed constant) and end up with a time-dependent speed for the oscillatory motion that will involve acceleration also.

If we study the time dependence in the previous equation, we can determine the acceleration to be:

a(t) = – ω2 · A · sin (θ) = –ω2 · A · sin (ω · t)

But we already have defined the elastic force, so how does this acceleration translate into the force? Imagine we extend the spring and then release it. The oscillations will occur due to the elastic force that tries to restore equilibrium. So in the absence of other forces, the elastic force is the net force:

F = m · a = Felastic

m · a = –k · Δx

m · [–ω2 · A · sin (w · t) ] = –k · Δx

If we consider that at equilibrium, the displacement is zero, then:

Δx = xx0 = x – 0 = x

And plugging this back into the forces equation, we have:

m · [–ω2 · A · sin (ω · t) ] = –k · x

Or:

m · [ –ω2 · A · sin (ω · t)] = –k · A · sin (ω · t)

Eliminating the similar factors from the left- and right-hand sides, we get:

k = m · ω2

And the elastic force is:

Felastic = – m · ω2 · A · sin (ω · t)

Felastic = –k · A · sin (ω · t)

Because the sine function varies between –1 and + 1, the elastic force is seen to be variable with time. Its extremes are:

Felastic MAX = + k · A

Felastic MIN = –k · A

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