Find a model for simple harmonic motion if the position at t equals 0 is 2 the amplitude is 2 centimeters and the period is 6 seconds?

Answer 1

I'm not entirely sure about this, but imagine that you have a coordinate system where the x-axis is in units of seconds (variable name t) and the y-axis is in units of centimeters (variable name d (for displacement)). Simple harmonic motion would indicate a simple periodic motion pattern, which would hint at a sine or cosine function. Using the above coordinate system, an unmodified cosine wave would have an amplitude of 1.0 cm and a period of 2(pi) seconds:

d=cos(t)

I chose cosine because cosine at t=0 (x=0) would be at whatever its amplitude is.

To get the amplitude to be 2.0 cm is easy, you must simply modify the amplitude of the cosine function by multiplying it by 2. So far:

d=2cos(t)

To modify the period to fit the problem statement, you must start with the fact that an unmodified cosine function would have a period of 2(pi). If you were to cut the period in half, you would multiply the argument inside the function by 2, which would give it a period of (pi). This would look like:

y=cos(2x)

So, if multiplying by 2 changes the period by 1/2, what would we need to multiply by to get a period of 6 seconds? 6 seconds is 6/2(pi)=3/(pi) of 2(pi), to get 3/(pi) of the current period, we must multiply the interior argument by (pi)/3. This gives us a final model of:

d=2cos((pi/3)t)

If this model is correct, one period of the cosine wave will complete by t=6 seconds. This means that the model will return values of 2 at t=0 and t=6. Let's check:

d(0)=2cos((pi/3)(0))=2cos(0)=2*1=2

d(6)=2cos((pi/3)(6))=2cos(2(pi))=2*1=2

These values check out. This means that our model works. So, the final answer is:

d=2cos((pi/3)t)