# how would I do this.....

Printable View

• Jun 5th 2007, 01:14 PM
kassandra
how would I do this.....
The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day? :confused:
• Jun 6th 2007, 06:46 AM
Pterid
Well, I would use this general result on 'combining errors' in things that are normally distributed:

If you have a variable $Z$ normally distributed about a mean value $\bar{Z}$ with variance $\sigma_{Z}^2$, and a function $F(Z)$, the function will be approximately normally distributed about $\bar{F} = F(\bar{Z})$with a variance

$\sigma_A^2 = \left( \frac{\partial F}{\partial Z} \right)^2 \sigma_Z^2$

In this case, the variable is the length $L$ with $\bar{L} = 10cm, \sigma_L^2 = (0.01cm)^2.$

The function is the period $T(L)$ given by:

$T = 2\pi \sqrt{\frac{L}{g}}$

so $\frac{\partial T}{\partial L} = \pi \sqrt{\frac{1}{gL}}.$

You need to find the standard deviation $\sigma_T$ where $\sigma_T^2 = \left( \frac{\partial T}{\partial L} \right)^2 \sigma_L^2 = \pi^2 \frac{1}{gL} \sigma_L^2$.

Use $L = \bar{L}$ in that expression.

...and then, using the properties of the normal distribution, you can find the probability of $T$ being wrong by a certain amount (enough to make the clock slow down by a minute a day). There's quite a bit of calculating left for you to do, but hopefully that should help you on your way.

- - -

You may have a different way to do it... it really depends on what you've been taught. :)