Finding Amplitude in Harmonic Function: Solving for Time t

  • Thread starter Thread starter Niles
  • Start date Start date
  • Tags Tags
    Amplitude
Niles
Messages
1,834
Reaction score
0

Homework Statement


Hi all.

I have the following harmonic function:

<br /> V(t)=A\cos(\omega t)\exp(-Ct),<br />


where C is a constant, and A is the amplitude. I need to find the time t, where the amplitude is A/2. This gives me:

<br /> V(t)=A\cos(\omega t)\exp(-Ct) = \frac{A}{2},<br />

but how do I solve this equation?

Thanks in advance.


Niles.
 
Physics news on Phys.org
Well, the obvious first step is to cancel the "A"s: cos(\omega t)e^{-Ct}= 1/2. Next, I think I would write the cosine in exponential form: cos(\omega t)= (e^{it}+ e^{-it})/2 so cos(\omega t)e^{-Ct}= (e^{(-C+ i\omega)t}+ e^{(C-i\omega)t})2= 1/2
 
Ahh, great.

If I was given a function on the form:

<br /> V(t)=(A\cos(\omega t)+B\sin(\omega t)\exp(-Ct),<br />

then writing the sines and cosines as exponentials would be the way to go too. But am I even correct to say that the time t when the amplitude of the oscillation of V(t) is half of the original amplitude is when V(t) = A/2, where A is the amplitude?
 
Last edited:
Yes, you said "find the time find the time t, where the amplitude is A/2". If the initial amplitude is A, then half of it is A/2.
 
Need numerical solve.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top