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  • Compare simple harmonic motion with uniform circular motion.

The figure shows a clock-wise rotating empty merry go round with iron bars holding the decorated wooden horse statues, four in each column.
The horses on this merry-go-round exhibit uniform circular motion. (credit: Wonderlane, Flickr)

There is an easy way to produce simple harmonic motion by using uniform circular motion. [link] shows one way of using this method. A ball is attached to a uniformly rotating vertical turntable, and its shadow is projected on the floor as shown. The shadow undergoes simple harmonic motion. Hooke’s law usually describes uniform circular motions ( ω size 12{ω} {} constant) rather than systems that have large visible displacements. So observing the projection of uniform circular motion, as in [link] , is often easier than observing a precise large-scale simple harmonic oscillator. If studied in sufficient depth, simple harmonic motion produced in this manner can give considerable insight into many aspects of oscillations and waves and is very useful mathematically. In our brief treatment, we shall indicate some of the major features of this relationship and how they might be useful.

The given figure shows a vertical turntable with four floor projecting light bulbs at the top. A smaller sized rectangular bar is attached to this turntable at the bottom half, with a circular knob attached to it. A red colored small ball is rolling along the boundary of this knob in angular direction, and the lights falling through this ball are ball making shadows just under the knob on the floor. The middle shadow is the brightest and starts fading as we look through to the cornered shadow.
The shadow of a ball rotating at constant angular velocity ω size 12{ω} {} on a turntable goes back and forth in precise simple harmonic motion.

[link] shows the basic relationship between uniform circular motion and simple harmonic motion. The point P travels around the circle at constant angular velocity ω size 12{ω} {} . The point P is analogous to an object on the merry-go-round. The projection of the position of P onto a fixed axis undergoes simple harmonic motion and is analogous to the shadow of the object. At the time shown in the figure, the projection has position x size 12{x} {} and moves to the left with velocity v size 12{v} {} . The velocity of the point P around the circle equals v ¯ max size 12{ {overline {v}} rSub { size 8{"max"} } } {} .The projection of v ¯ max size 12{ {overline {v}} rSub { size 8{"max"} } } {} on the x size 12{x} {} -axis is the velocity v size 12{v} {} of the simple harmonic motion along the x size 12{x} {} -axis.

The figure shows a point P moving through the circumference of a circle in an angular way with angular velocity omega. The diameter is projected along the x axis, with point P making an angle theta at the centre of the circle. A point along the diameter shows the projection of the point P with a dotted perpendicular line from P to this point, the projection of the point is given as v along the circle and its velocity v subscript max, over the top of the projection arrow in an upward left direction.
A point P moving on a circular path with a constant angular velocity ω size 12{ω} {} is undergoing uniform circular motion. Its projection on the x-axis undergoes simple harmonic motion. Also shown is the velocity of this point around the circle, v ¯ max size 12{ {overline {v}} rSub { size 8{"max"} } } {} , and its projection, which is v size 12{v} {} . Note that these velocities form a similar triangle to the displacement triangle.

To see that the projection undergoes simple harmonic motion, note that its position x size 12{x} {} is given by

x = X cos θ , size 12{x=X"cos"θ","} {}

where θ = ω t size 12{θ=ω`t} {} , ω size 12{ω} {} is the constant angular velocity, and X size 12{X} {} is the radius of the circular path. Thus,

x = X cos ω t . size 12{x=X"cos"ω`t} {}

The angular velocity ω size 12{ω} {} is in radians per unit time; in this case size 12{2π} {} radians is the time for one revolution T size 12{T} {} . That is, ω = / T size 12{ω=2π/T} {} . Substituting this expression for ω size 12{ω} {} , we see that the position x size 12{x} {} is given by:

x ( t ) = cos t T . size 12{x \( t \) ="cos" left ( { {2π`t} over {T} } right )} {}

This expression is the same one we had for the position of a simple harmonic oscillator in Simple Harmonic Motion: A Special Periodic Motion . If we make a graph of position versus time as in [link] , we see again the wavelike character (typical of simple harmonic motion) of the projection of uniform circular motion onto the x size 12{x} {} -axis.

The given figure shows a vertical turntable with four floor projecting light bulbs at the top. A smaller sized rectangular bar is attached to this turntable at the bottom half, with a circular knob attached to it. A red colored small ball is rolling along the boundary of this knob in angular direction. The turnaround table is put upon a roller paper sheet, on which the simple harmonic motion is measured, which is shown here in oscillating waves on the paper sheet in front of the table. A graph of amplitude versus time is also given alongside the figure.
The position of the projection of uniform circular motion performs simple harmonic motion, as this wavelike graph of x size 12{x} {} versus t size 12{x} {} indicates.

Now let us use [link] to do some further analysis of uniform circular motion as it relates to simple harmonic motion. The triangle formed by the velocities in the figure and the triangle formed by the displacements ( X , x , size 12{X,x,} {} and X 2 x 2 size 12{ sqrt {X rSup { size 8{2} } - x rSup { size 8{2} } } } {} ) are similar right triangles. Taking ratios of similar sides, we see that

v v max = X 2 x 2 X = 1 x 2 X 2 . size 12{ { {v} over {v rSub { size 8{"max"} } } } = { { sqrt {X rSup { size 8{2} } - x rSup { size 8{2} } } } over {X} } = sqrt {1 - { {x rSup { size 8{2} } } over {X rSup { size 8{2} } } } } } {}

We can solve this equation for the speed v size 12{v} {} or

v = v max 1 x 2 X 2 . size 12{v=v rSub { size 8{"max"} } sqrt {1 - { {x rSup { size 8{2} } } over {X rSup { size 8{2} } } } } } {}

This expression for the speed of a simple harmonic oscillator is exactly the same as the equation obtained from conservation of energy considerations in Energy and the Simple Harmonic Oscillator .You can begin to see that it is possible to get all of the characteristics of simple harmonic motion from an analysis of the projection of uniform circular motion.

Finally, let us consider the period T size 12{T} {} of the motion of the projection. This period is the time it takes the point P to complete one revolution. That time is the circumference of the circle X size 12{2πX} {} divided by the velocity around the circle, v max size 12{v rSub { size 8{"max"} } } {} . Thus, the period T size 12{T} {} is

T = 2πX v max . size 12{T= { {2πX} over {v rSub { size 8{"max"} } } } } {}

We know from conservation of energy considerations that

v max = k m X . size 12{v rSub { size 8{"max"} } = sqrt { { {k} over {m} } } X} {}

Solving this equation for X / v max size 12{X/v rSub { size 8{"max"} } } {} gives

X v max = m k . size 12{ { {X} over {v rSub { size 8{"max"} } } } = sqrt { { {m} over {k} } } } {}

Substituting this expression into the equation for T size 12{T} {} yields

T = m k . size 12{T=2π sqrt { { {m} over {k} } } "."} {}

Thus, the period of the motion is the same as for a simple harmonic oscillator. We have determined the period for any simple harmonic oscillator using the relationship between uniform circular motion and simple harmonic motion.

Some modules occasionally refer to the connection between uniform circular motion and simple harmonic motion. Moreover, if you carry your study of physics and its applications to greater depths, you will find this relationship useful. It can, for example, help to analyze how waves add when they are superimposed.

Identify an object that undergoes uniform circular motion. Describe how you could trace the simple harmonic motion of this object as a wave.

A record player undergoes uniform circular motion. You could attach dowel rod to one point on the outside edge of the turntable and attach a pen to the other end of the dowel. As the record player turns, the pen will move. You can drag a long piece of paper under the pen, capturing its motion as a wave.

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Section summary

A projection of uniform circular motion undergoes simple harmonic oscillation.

Problems&Exercises

(a)What is the maximum velocity of an 85.0-kg person bouncing on a bathroom scale having a force constant of 1 . 50 × 10 6 N/m size 12{1 "." "50" times "10" rSup { size 8{5} } "N/m"} {} , if the amplitude of the bounce is 0.200 cm? (b)What is the maximum energy stored in the spring?

a). 0.266 m/s

b). 3.00 J

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A novelty clock has a 0.0100-kg mass object bouncing on a spring that has a force constant of 1.25 N/m. What is the maximum velocity of the object if the object bounces 3.00 cm above and below its equilibrium position? (b) How many joules of kinetic energy does the object have at its maximum velocity?

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At what positions is the speed of a simple harmonic oscillator half its maximum? That is, what values of x / X size 12{x/X} {} give v = ± v max / 2 size 12{v= +- v rSub { size 8{"max"} } /2} {} , where X size 12{X} {} is the amplitude of the motion?

± 3 2 size 12{ +- { { sqrt {3} } over {2} } } {}

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A ladybug sits 12.0 cm from the center of a Beatles music album spinning at 33.33 rpm. What is the maximum velocity of its shadow on the wall behind the turntable, if illuminated parallel to the record by the parallel rays of the setting Sun?

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Questions & Answers

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Dimensional Analysis. The study of relationships between physical quantities with the help of their dimensions and units of measurements is called dimensional analysis. We use dimensional analysis in order to convert a unit from one form to another.
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states that electric current in a given metallic conductor is directly proportional to the potential difference applied between its end, provided that the temperature of the conductor and other physical factors such as length and cross-sectional area remains constant. mathematically V=IR
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Gundala
A body travelling at a velocity of 30ms^-1 in a straight line is brought to rest by application of brakes. if it covers a distance of 100m during this period, find the retardation.
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v^2-u^2=2as v=0,u=30,s=100 -30^2=2a*100 -900=200a a=-900/200 a=-4.5m/s^2
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The change in position of an object with respect to time
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mass × acceleration OR Work done ÷ distance
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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