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Simple harmonic motion

  • Thread starter denian
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  • #1
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a particle of mass m moving in such manner that its displacement x varies with time t following equation x = a sin wt. ( w = omega )
where a and w are constants.
what is the average KE for the motion of the particle in one rev?

isnt the answer suppose to be 1/2 mw(sq)a(sq) ?
 

Answers and Replies

  • #2
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In my physics class, to solve a problem like that, we used the equation KE=(1/2)m*v^2. So, when given the equation for x, take the derivative to get velocity, and insert it into the equation.
 
  • #3
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yeah, that is how they derive the equation
1/2 mw^2 a^2.

i use that way, but the answer is 1/4mw^2 a^2
 
  • #4
HallsofIvy
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The problem asked for the average kinetic energy.

Given a continuous function, its "average value" on an interval from a to b is [tex]\frac{\int_a^b f(x)dx}{b-a} [/tex]

For this problem, the kinetic energy is (1/2)mv2= (1/2)ma2w2cos2(wt). One revolution will occur as wt goes from 0 to 2π so as t goes from 0 to 2π/w and so the integration is from 0 to 2π/w.

The integral itself is ma2w2(π/w) but then we need to divide by 2&pi/w-0: the average kinetic energy is (m/2)a2w.
 

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