Varaition!!

corpse_bride · November 24th 2007, 10:15 PM

OK, I have a question here that's part of a homework task I have due tomorrow:

The time taken, t, for a pendulum to swing varies as the sqare root of its length, L. If one swing of a pendulum 98cm long takes 2 seconds, find the time taken for one swing of a pendulum 32cm long.

I know of varaition, direct variation etc, but what do I do in this case? I can't work it out.

Any help is greatly appreciated

**Soroban** · November 24th 2007, 10:41 PM

Hello, corpse_bride!

The time $t$ for a pendulum to swing varies as the sqare root of its length $L.$
If one swing of a pendulum 98cm long takes 2 seconds,
find the time taken for one swing of a pendulum 32cm long.

This is direct variation . . .

Write the equation! . $t \;=\;k\sqrt{L}$

We are told: when $L = 98,\;t = 2$
. . So we have: . $2 \;=\;k\sqrt{98}\quad\Rightarrow\quad k \:=\:\frac{2}{\sqrt{98}} \:=\:\frac{2}{7\sqrt{2}} \:=\:\frac{\sqrt{2}}{7}$
Hence: . $t \;=\;\frac{\sqrt{2}}{7}\sqrt{L}$

Now find $t$ when $L = 32.$

**kalagota** · November 24th 2007, 10:45 PM

Originally Posted by corpse_bride

OK, I have a question here that's part of a homework task I have due tomorrow:

The time taken, t, for a pendulum to swing varies as the sqare root of its length, L. If one swing of a pendulum 98cm long takes 2 seconds, find the time taken for one swing of a pendulum 32cm long.

I know of varaition, direct variation etc, but what do I do in this case? I can't work it out.

Any help is greatly appreciated

we have $t = k\sqrt L$ from the given, we have $2 = k \sqrt {98}$ or $k = \frac{2}{\sqrt {98}}$

using this for the second pendulum, we have $t = \frac{2\sqrt {32}}{\sqrt {98}} = 2 \sqrt{\frac{32}{98}} = (2) \left( {\frac{4}{7}} \right) \left( \sqrt 1 \right) = \frac{8}{7} \approx 1.142857.$

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