Damped oscillations - Solution

To find the position of a peak we use the fact that the cosine function is a maximum when its argument is 0. So we solve:
10*pi*t+pi/4 = 0

Rearranging gives t=-1/40.

As this is a negative value we add on the value of the period to find the first peak in the positive time zone. The period is given by 2*pi / 10*pi =0.2, so the first peak will occur at:

t=-1/40+0.2 = 0.175

We use the General Function to plot (e^(-0.5*t))*cos(10*pi*t+pi/4)
And the Display Position to enter t=0.166667 and get the following result for the first peak

Considering any two second interval

To confirm this result for any 2 second interval, we consider the exponential part of the function, e^(-0.5*t), and two values t0 and t0+2.

• At time t0 we have e^(-0.5*t0)
  
• At time t0+2 we have e^(-0.5*(t0+2))
  
• The proportional decrease found is: e^(-0.5*(t0+2)) / e^(-0.5*t0)

We use properties of exponentials to simplify this to: e^(-0.5*(t0+2)+0.5*t0) = e^(-0.5*2) = e^-1 = 1/e
= 0.36788 to 6 s.f.

This confirms our result from above and we have established that the oscillations will decrease by a factor of 0.36788 in any 2 second interval.

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