# Math Help - Pendulum problem

1. ## Pendulum problem

A pendulum of period 2s is released from rest at an inclination of 5 degrees to the downward vertical. What is its angular velocity when it reaches the vertical?

$\tau=\frac{2\pi}{\omega}$
$\omega=\frac{2\pi}{2}$
$\omega=\pi$

however the answer in book is 15.7degrees/sec

Further,
When it first returns to the starting point, it is given an impulsive blow towards the vertical that increases the amplitude of swing to 10degrees. Find its subsequent angular postion as a function of time.

the pendulum has a restoring force of :
$F=-mgsin\theta$
$m\ddot{x}=-mgsin\theta$
i realise the acceleration is centripetal so $\ddot{x}=r(\omega)^2$
also i think you should go down the the integration with respect to time and use angular measurements instead of linear. This is where im stuck.

2. Originally Posted by FGT12
A pendulum of period 2s is released from rest at an inclination of 5 degrees to the downward vertical. What is its angular velocity when it reaches the vertical?

$\tau=\frac{2\pi}{\omega}$
$\omega=\frac{2\pi}{2}$
$\omega=\pi$

however the answer in book is 15.7degrees/sec
The book is correct.

The small amplitude period of a pendulum of length $L$ is:

$\tau = 2 \pi \sqrt{\dfrac{L}{g}}$

Use this and the given period to find the length of your pendulum.

Now use conservation of energy to equate the change in potential energy between the highest point and the lowest point of the swing to determine the speed of the bob at the lowest point and use that to calculate its angular velocity.

CB

3. so what does the $\omega$ stand for in the formula $\tau=\frac{2\pi}{\omega}$. I thought it was angular velocity?

4. It is.

5. so why is my line of working incorrect? I noticed it was the recipbrocal of the correct answer...

6. Originally Posted by FGT12
so why is my line of working incorrect? I noticed it was the recipbrocal of the correct answer...
Is the pendulum completing 360 degree rotation at a period of 2s? No, what you have is nonsense unrelated to small amplitude pendulum oscillation

CB

CB

7. I still dont see how the second part works

8. Originally Posted by FGT12
I still dont see how the second part works
After the impulse the period will be the same (remember for small oscillations the period is independent of amplitude) and the amplitude will be 10 degrees, but there will be a phase term:

$\theta(t)=10 \sin\left(\frac{2 \pi t}{T} + \phi\right)$ degree

Take the time of the impulse as zero then $\theta(0)=5$ degree and solve for $\phi$

CB

9. Originally Posted by FGT12
A pendulum of period 2s is released from rest at an inclination of 5 degrees to the downward vertical. What is its angular velocity when it reaches the vertical?...

... the answer in book is 15.7degrees/sec
For the 'energy conservation law' when the inclination of the pendulum reaches the vertical, the loss of potential energy must be equal to the 'cinetic energy' and that means that, if we indicate with $\omega_{0}$ the 'maximum angular speed' , it must be...

$\displaystyle \frac{l}{2}\ m\ \omega_{0}^{2} = l\ m\ g\ (1-\cos \theta) \rightarrow \omega_{0} = \sqrt{2\ g\ (1-\cos \theta)$ (1)

For $\theta$ = 5 degrees and g = 9.8 meters/sec^2 the (1) gives $\omega_{0}$ = .27310056... radians/sec = 15.647509... degrees/sec. It is remarkable the fact that the result does't depend from l, m and from the period of the pendulum...

Kind regards

$\chi$ $\sigma$

10. In my textbook it shows the answer as
$\theta = 5^\circ cos(\pi t) - 8.66^\circ sin(\pi t)$#
It is this i cannot see how to get

11. Originally Posted by FGT12
In my textbook it shows the answer as
$\theta = 5^\circ cos(\pi t) - 8.66^\circ sin(\pi t)$#
It is this i cannot see how to get
Try applying some trig identites to my previous post (after changing the sine to a cosine if you prefer, it makes no difference in principle) and then fitting to the data you are given.

CB