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Old Apr 10th 2015, 07:23 PM   #1
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Question Particle executing simple harmonic motion

This refers to question 14. in the scan I attached. A particle executes simple harmonic motion with amplitude of 3.0 cm. At what displacement from the midpoint of its motion does its speed equal one half of its maximum speed? The answer to the problem is 2.6 cm. I don't know how to obtain this answer. I have included my work. Thanks.
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Old Apr 11th 2015, 02:30 AM   #2
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You are on the right track, but getting bogged down in the equations

Taking the displacement motion as a cosine function, then (as you correctly indicate) the speed will be given by a sine function.

Ignore scaling factors (for now) you want to know when the sine function is at half its maximum.
The maximum for an unscaled sine function is 1.0 so we want to know when the sine function is 0.5.

The inverse sine of 0.5 is 30 (degrees),
what is the displacement (cosine function) at 30 (degrees)?
(don't forget to bring back the scaling factor at this point)

Note that I have been a little bit lazy,
I haven't used omega t or phi or even radians,
but the answer is the same,
the rest of the equation is (in this example) superfluous (for other problems they may be crucial).

I guess it is a bit of a knack identifying the important features of the equations for a given problem.
I can definitely remember floundering in almost exactly the same manner as you during my school days.
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Last edited by MBW; Apr 11th 2015 at 02:33 AM.
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Old Apr 11th 2015, 03:10 AM   #3
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I've been thinking a bit more about my answer and realised I could explain a bit more about why the scaling factors etc. can be ignored.

firstly the 3 cm.
The measuring unit is actually quite arbitrary, we could use inches or parsecs or...
I choose to use quogs, which I have just invented for the occasion.
one quog happens to be exactly 3 cm.
This has the advantage that the the scaling factor becomes 1 and so become trivial and disappears from the problem.

Similarly the omega in the equation is scaling factor between time and and the wavelength of the motion.
choose ticks (of just the right length) instead of seconds and omega disappears.
(note that the use of degrees or radians is also just a scaling factor, I chose to use degrees in my answer simply because that is the default setting on my calculator).

Finally, phi is an offset, which indicates when you started timing with respect to the cycle of the motion.
In a real experiment this would be important, but the advantage of a thought experiment is that you can start your stopwatch whenever you like.
If you start your stopwatch at precisely the start of the cycle, then phi will be zero and disappears from the problem.

so work out the problem using ticks and quogs with a perfect click of the stopwatch.
Only at the end do you need to re-scale your answer from quogs to centimetres.
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