Solving exponential simultaneous equations

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SUMMARY

The discussion focuses on solving exponential simultaneous equations derived from a sinusoidal waveform described by the equation T = Ae-Ktsin(wt + ø). The values for w and ø are established as w = 40π (or 125.66) and ø = -1.885. The main challenge is to find the constants A and K using two simultaneous equations derived from peak values. The correct approach involves recognizing the need for a negative exponent when dividing the equations, ultimately leading to K = ln(2.875)/0.05 = 21.121, which allows for a consistent value of A.

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MathsDude69
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Homework Statement



The actual problem shows a graph however I can state all the information. The graph is of a sinusiodal waveform where the amplitude is decaying exponentially. The formula for the graph is given by the equation:

T = Ae-Ktsin(wt + ø)

The question is to find A,K,w and ø

Being quite confident in sinusoidal waveforms I can tell you that:

w = 40 x pi or 125.66 (whichever tickles your fancy)
ø = -1.885

However I am stuck with the A and K.

Assuming that the maximum peaks occur when sin(wt + ø) = 1 then:

0.23 = Ae-K0.0275

0.08 = Ae-K0.0775

I now have 2 points to solve simultaneously for A and K.

The Attempt at a Solution



0.23/0.08 = Ae-K0.0275/Ae-K0.0775

2.875 = e-K0.5

K = (1/0.5) x ln(2.875) = 2.1121


When you plug this back into the two equations however you get two different answers for A and A is supposed to be a constant. Can anyone see where I am goign wrong here?

Thanks in advance for any help.
 
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0.23/0.08 = Ae-K0.0275/Ae-K0.0775

2.875 = e-K0.5
K = (1/0.5) x ln(2.875) = 2.1121
Oops! That should be e0.05k in the second line. When dividing two powers you subtract the exponents. Also, if we pretend that the 2nd line was right, then the 3rd line is missing a negative in front of the fraction.

But anyway, there should be no negative:

2.875 = e^{0.05k}

k = \frac{ln(2.875)}{0.05} = 21.121

Now you should get a single value for A.


01
 

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