# Damped Simple Harmonic Motion

## Homework Statement

http://img534.imageshack.us/img534/6503/questiono.jpg [Broken]

## The Attempt at a Solution

(a) I'm not sure how to approach this part. I appreciate any help to get me started.

(b) I know that the period of oscillation is given by

$T'=\frac{2 \pi}{\omega '}$

I can use this later on to express my results in terms of the period T'. The one term in the given expression for x(t) that can turn negative as time increases would be the δ in the cosine. So how do I evaluate the right value for the phase delta?

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for first one ,differentiate x(t) wrt time and put velocity so obtained=0 at t=0,perhaps that will do.

for first one ,differentiate x(t) wrt time and put velocity so obtained=0 at t=0,perhaps that will do.

Thanks. I'm still a bit confused. When I differentiated the expression with respect to t I got:

$x'(t)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) e^{-t/\tau} (\tau \omega ' \sin(\omega' t + \delta)+\cos(\omega' t + \delta))}{\tau}$

I've found x'(0):

$x'(0)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) (\tau \omega ' \sin( \delta)+\cos(\delta))}{\tau}$

But I have to show that δ is given by:

$\tan \delta = - \frac{1}{\omega ' \tau}$

So I'm not sure, how can do I use the expression for x'(0) to show this?

Did you mean I write down the velocity as:

$v(t)=\omega ' . \frac{A_0}{\cos (\delta)} e^{-t/\tau} \cos (\omega ' t + \delta)$

$\therefore \ v(0) =A_0 \omega '$

So, how does this help? Thanks. I'm still a bit confused. When I differentiated the expression with respect to t I got:

$x'(t)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) e^{-t/\tau} (\tau \omega ' \sin(\omega' t + \delta)+\cos(\omega' t + \delta))}{\tau}$

I've found x'(0):

$x'(0)=\frac{\left( \frac{A_0}{\cos (\delta)}\right) (\tau \omega ' \sin( \delta)+\cos(\delta))}{\tau}$

But I have to show that δ is given by:

$\tan \delta = - \frac{1}{\omega ' \tau}$

So I'm not sure, how can do I use the expression for x'(0) to show this?

ok.you have done it.equate it to zero(because velocity is initially zero) don't you get what is required.

ok.you have done it.equate it to zero(because velocity is initially zero) don't you get what is required.

I'm a bit confused. What do I have to do once I equate that expression to zero?

by equating you will get,
w'(tau)sin(delta)=-cos(delta)
which gives
tan(delta)=-1/w'(tau)

Oh oops I didn't see that part. Thank you so much for the help I really appreciate that!!

Any ideas how to approach the second part?

the best way to it will be to draw(or see elsewhere) the graph and judge when x(t) will be zero for first time or just put it equal to zero and find t for first time.the graph is cosine curve ,shifted leftward and with decreasing amplitude.max. negative displacement will occur after first cycle means after when it crosses first zero and see wikipedia for other assistance.

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If I want to calculate it. We have by setting it equal to zero:

x(0) = (A0/cos) cos (δ) = 0

So what value do I use for δ? Do I have to use tan δ = 1/ω'τ and then substitute ω'=2π/T'?

I am not saying to put t=0 and then equate x=0.you have to find t by equating it.since
e(-t/tau) can not be zero.cosine part will be zero find when for the first time this will be zero.
cos(w't+delta)=0,and choose cos(pi/2) from here you will get t for first zero.max. negative displacement can be obtained at T'/2 time from the beginning means after
-delta/w'.can you tell the answer so that I can check it.