Solving a Simple Pendulum Problem with Math

[Lord Anoobis]

Homework Statement

A 100g mass on a 1.0m long string is pulled 8.0 degrees to one side and released. How long does it take for the pendulum to reach 4.0 degrees on the opposite side?

Homework Equations

$T = 2\pi \sqrt\frac{L}{g}$
$x(t) = A\cos\omega t$

The Attempt at a Solution

From the simple pendulum we get $\omega = \sqrt\frac{g}{L}$ which leads to:

$x(t) = A\cos\sqrt g t$

$A = 8.0$ and $x(t) = 4.0$ can be substituted directly since it results in the cosine of an angle then:

$\cos\sqrt g t = -0.50$

$t\sqrt g = \frac{4\pi}{3}$

$t = 1.3s$

Which happens to be twice the actual answer. The only possibility for error I see is $\arccos(-0.5)$.
What I don't see is why. My reasoning was that since we're looking at 4.0 degrees on the opposite side, the angle must be the second point where cosine is negative, ie. $\frac{4\pi}{3}$. What am I missing here?

 

[PeroK]

You need to be careful working in degrees.

I think you've identified the error. How did you evaluate the inverse cosine?

 

[Lord Anoobis]

Why are you working in degrees?

Because in this case it makes no difference since the division results in a dimensionless figure.

 

[PeroK]

Because in this case it makes no difference since the division results in a dimensionless figure.

Yes, but I'd be careful. It's an easy way to go wrong. If you draw a graph of cos, you'll see where you've gone wrong.

 

[Lord Anoobis]

You need to be careful working in degrees.

I think you've identified the error. How did you evaluate the inverse cosine?

There are two possibilities for $-0.5$, either $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$. I took the first figure to equate to 4.0 degrees on the side on which the swing started.

 

[PeroK]

There are two possibilities for $-0.5$, either $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$. I took the first figure to equate to 4.0 degrees on the side on which the swing started.

Wouldn't that be where $cos = 0.5$?

 

[Lord Anoobis]

Wouldn't that be where $cos = 0.5$?

I'm sure that's not the case.

 

[PeroK]

I'm sure that's not the case.

Are you sure you're sure?

Can you say where the pendulum is when?

$cos(\omega t) = 1$
$cos(\omega t) = 0.5$
$cos(\omega t) = 0$
$cos(\omega t) = -0.5$
$cos(\omega t) = -1$

(All for the first time)

 

[Lord Anoobis]

Are you sure you're sure?

Can you say where the pendulum is when?

$cos(\omega t) = 1$
$cos(\omega t) = 0.5$
$cos(\omega t) = 0$
$cos(\omega t) = -0.5$
$cos(\omega t) = -1$

(All for the first time)

I just noticed there's a typo in my initial post. It's supposed to be $x(t) = -4.0$ That's where the negative sign comes from.

 

[Lord Anoobis]

Are you sure you're sure?

Can you say where the pendulum is when?

$cos(\omega t) = 1$
$cos(\omega t) = 0.5$
$cos(\omega t) = 0$
$cos(\omega t) = -0.5$
$cos(\omega t) = -1$

(All for the first time)

I see what you mean here. The pendulum in question hasn't even completed one half of a full cycle. $\frac{4\pi}{3}$ is when it reaches 4.0 for the second time.