Question

In: Physics

A "seconds" pendulum is one that goes through its equilibrium position once each second. (The period...

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? (Give your answer to at least 4 decimal places.) Cambridge /Tokyo =

Expert Solution

First of all free fall accelerations are nothing but the acceleration due to gravity at that place.

Hence we are supposed to find the ratio of acceleration due to gravity at Cambridge to acceleration due to gravity at Tokyo.

Let gC and gT be the acceleration due to gravity at Cambridge and Tokyo respectively and lC and lT be the length of the simple pendulum at Cambridge and Tokyo respectively.

Given,

lC = 0.9942 m

lT = 0.9927 m

Time period of a simple pendulum is given by,

where,

T = Time period of simple pendulum (s)

l = Length of simple pendulum (m)

g = acceleration due to gravity at the place (m/s2)

Since it is a seconds pendulum, Time period at both the locations will be the same, ie, 2s

ie, TC = TT

ie,

Therefore,

genius_generous answered 2 days ago

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