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Physics/Simple Harmonic Motion

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Question
A hovering fair-sized insect rises a little during the downstroke of its wings and essentially free-falls during the upstroke.  The end result is that the creature oscillates in midair. Assume it typically falls about .212 mm per cycle. What is the wingbeat period?

Answer
If we assume that the insect falls 0.212mm=0.000212m using kinematics the time to fall can be determined from:
Df=1/2*a*t^2
0.000212=9.8/2*t-^2
Solving for t:
t=sqrt(0.000212/4.9)=0.00658sec
If we assume that approximately the same time is required to return the insect to the initial height we would need to double this time to find the period:
T=2*t=2*--0.00658sec=0.0131sec
By the way, in any case this is NOT simple harmonic motion. For it to be so there would have to be an equilibrium position, which there is not, and the restoring force from equilibrium would need to be proportional to the displacement from equilibrium. According to this problems statement the force accelerating the bug downward is gravity which is constant.

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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