derive the expression for decaying amplitude
Answers
The green line is $A(t) ~=~ e^{-t}$. It is the envelope of the oscillation. Obviously depending on the rate of decay of the amplitude, and the frequency, you'll get a different picture. But qualitatively you'll see an oscillating function whose amplitude decays away to zero. This should describe weak damping. We don't expect this to work too well in molasses. To get a more quantitative understanding we'll have to do some more math.
We'll try sticking $x(t) ~=~ A e^{\lambda t}$ into eqn. 1.60. Here again, $A$ is just a constant. We already differentiated this function before in eqns. 1.24 and 1.25 so we don't have to do it again. So we\begin{displaymath}
m\lambda^2 x + b \lambda x + kx ~=~ 0
\end{displaymath} (1.63)
Canceling the $x$'s
\begin{displaymath}
m\lambda^2 + b \lambda + k ~=~ 0
\end{displaymath} (1.64)
This is a quadratic equation for $\lambda$. Let's solve it:
\begin{displaymath}
\lambda ~=~ {-b \pm \sqrt{b^2 -4mk}\over 2m}
\end{displaymath} (1.65)
So we have two possible solutions for $\lambda$! They both solve the equation, and we have to have more information to figure out what to do with them. But for the moment, let's look at this equation more closely.
If the damping, $b$, is large, then the square root is real. However if $b^2 ~<~ 4mk$, then it becomes imaginary. have
Answer: put everything below on a horizontal line
Explanation:
−
2m
b
τ=−1⟹τ=
b
2m