Simple Harmonic Motion and free-fall accelerations

1. Apr 19, 2005

imnotsmart

A "seconds" pendulum is one that goes through its equilibrium position once each second. (The period of the pendulum is 2.000 s.) The length of a seconds pendulum is 0.9927 m at Tokyo and 0.9942 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
Cambridge / Tokyo) = 1 +?

How do I work this problem? Any help would be appreciated.

2. Apr 19, 2005

ramollari

As you observe, in Cambridge the value of g is higher than in Tokyo (g_C/g_T = 1 + ?).

Knowing that the periods are equal, then:

$$\frac{T_C}{T_T} = 1$$

or

$$\frac{2\pi \sqrt{\frac{l_C}{g_C}}}{2\pi \sqrt{\frac{l_T}{g_T}}} = 1$$

You can go on to find g_C/g_T in terms of the ratio l_C/l_T.

3. Apr 19, 2005

theCandyman

Are you familiar with this formula?

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

$$T$$ is the peroid, so set it equal to two and solve for the value $$g$$ with your given $$L$$ value.

4. Apr 19, 2005

ramollari

It is neater to express the ratio of free fall accelerations in terms of the ratio of the 1-sec pendulum lengths, than finding the free fall acceleration in each location and then computing the ratio.

Indeed the conclusion will be that:

$$\frac{g_C}{g_T} = \frac{l_C}{l_T} = \frac{0.9942}{0.9927} = 1 + 1.51 * 10^{-3}$$

5. Apr 19, 2005

theCandyman

Yes, yours is a much better way to compute it. I dislike dealing with ratios throught out a whole problem so I attempted it a different way.