# 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....

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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### 2 Answers

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,

l=0.9943m.

pi=3.1416.

T=2 sec.

solving the above equation for'g',we get the value of 'g' as 9.81335 m/s2 which is option (1).