Question

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9923 m at Tokyo and 0.9941 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

Answer 1

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998

Step-by-step explanation:

We know the equation

`T=2\pi \sqrt{(l)/(g)}`

where l is length of pendulum, g is acceleration due to gravity and T is period.

Rearranging

`g= (4\pi^2l)/(T^2)`

Length of pendulum in Tokyo = 0.9923 m

Length of pendulum in Cambridge = 0.9941 m

Period of pendulum in Tokyo = Period of pendulum in Cambridge = 2s

We have

`\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}= \frac{\frac{4\pi^2 l_{\texttt{Tokyo}}}{ T_{\texttt{Tokyo}}^2}}{\frac{4\pi^2 l_{\texttt{Cambridge}}}{ T_{\texttt{Cambridge}}^2}}\\\\\frac{ g_{\texttt{Tokyo}}}{ g_{\texttt{Cambridge}}}=((0.9923)/(2^2))/((0.9941)/(2^2))=0.998`

Ratio of free fall acceleration of Tokyo to Cambridge = 0.998