In Cambridge, England, a seconds pendulum is 0.9943 m long. What is the free-fall acceleration in Cambridge?
1. 9.81335 m/s^2
2. 9.81 m/s^2
3. None of these
4. 9.78366 m/s^2
5. 9.81441 m/s^2
6. 9.78473 m/s^2
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A seconds pendulum is one that takes one second to swing to an extreme position and return to the equilibrium position. This gives the time period of a seconds pendulum as 2 seconds. The time period of any pendulum of length L is equal to 2*pi*sqrt(L/g) where g is the free fall acceleration.
As the length of the seconds pendulum in Cambridge is 0.9943 m, it gives:
2 = 2*pi*sqrt(L/g)
=> sqrt(L/g) = 1/pi
=> L/g = 1/pi^2
=> g = L*pi^2
=> g = 9.813347 m/s^2
The correct answer is option 1, 9.81335 m/s^2
the answer is 9.81335 m/s2.
the formula for the time period of a simple pendulum is T=2 pi square root of (l/g).
using the given data,
solving the above equation for'g',we get the value of 'g' as 9.81335 m/s2 which is option (1).
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