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16.6 Uniform circular motion and simple harmonic motion  (Page 2/2)

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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 2π 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 = 2π 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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Test prep for ap courses

In the equation x   =   A   sin   w t, what values can the position x take?

  1. −1 to +1
  2. – A to + A
  3. 0
  4. – t to t
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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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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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