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Unknown008 · Jan 16, 2010, 12:50 AM

I'm having trouble understanding that question. And as such, cannot answer it:

Qu:
A point moves with S.H.M. along an x-axis according to the equation:

$\frac{d^2x}{dt^2} + Ax = 0$

The period of this motion is:

A. $\frac{\sqrt A}{2\pi}$

B. $\frac{\sqrt A}{\pi}$

C. $\frac{\pi}{\sqrt A}$

D. $\frac{2\pi}{\sqrt A}$

E. $\frac{\pi}{2\sqrt A}$

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It is that differentiation of second order which is bothering me. I don't understand what that means here :( I want to know what to do with the equation. I tried to integrate but I don't think that's how to do it.

ebaines · Jan 16, 2010, 04:27 PM

Put in the equation

$x = Csin(\omega t)$ ,

and see what happens - it's second derivative is: $\frac {d^2x} { dt^2 }= -C \omega^2sin(wt)$

From this you should be able to see what the value of $\omega$ is. The period of motion is then: $T = \frac {2 \pi } {\omega}$

Unknown008 · Jan 16, 2010, 09:43 PM

I'm not sure of it at all (especially about the x and the constant C), but here I go:

$\frac{d^2x}{dt^2} = -C \omega ^2 sin(\omega t)$

Sub in from the given equation:

$-Ax = -C \omega ^2 sin(\omega t)$

$Ax = C \omega ^2 sin(\omega t)$

At max amplitude, sin becomes 1

$Ax = C \omega ^2$

$\omega = \sqrt{\frac{Ax}{C}}$

Assuming x = C;

$\omega = \sqrt{A}$

Hence answer is D.

Is that the way?

ebaines · Jan 17, 2010, 10:03 AM

Almost right... you have the right answer, but made a small slip along the way. In the second equation, where you have

$<br /> Ax = C \omega ^2 sin(\omega t)<br />$

remember that $x$ is a function of time: $x= C sin(\omega t)$ , so:

$<br /> Ax = A C sin(\omega t) = C \omega^2 sin( \omega t)<br />$

Therefore, $A = \omega ^2$ . This leads to the answer being D.

Unknown008 · Jan 17, 2010, 10:05 AM

Ah, many thanks! I would have never gone through that :)

I may be posting more physics multiple choice questions in the following days. Cya around! :)

ebaines · Jan 17, 2010, 10:13 AM

You're welcome! By the way, a more formal approach would have been to use $x = C sin(\omega t + \phi)$ , where $\phi$ represents the phase angle at t = 0. This may come in handy for you later on if you get into problems that require taking into account the initial boundary conditions. For this problem it didn't matter, as they were asking only for the period of motion and didn't provide boundary conditions, so we could assume that $\phi =0$ .

Unknown008 · Jan 17, 2010, 10:17 AM

I'm taking that boundary conditions are phase differences?

Thanks again! :)