A pendulum that moves through its equilibrium position once every 1.000 s is sometimes called a seconds pendulum.

(a) What is the period of any seconds pendulum?

(b) In Cambridge, England, a seconds pendulum is 0.9942 m long. What is the free-fall acceleration in Cambridge? (c) In Tokyo, Japan, a seconds pendulum is 0.9927 m long. What is the free-fall acceleration in Tokyo?

a) 2 s

b) 9.8122 m/s²

c) 9.7976 m/s²

Explanation:

First, let's do logic here to solve every question.

a) If the pendulum moves through it's equilibrium position once every 1.000s or one second only, this means that the period it's the double of time, ergo, 2.000 s. This is because a pendulum goes from right to the left and passes through it's equlibirum twice, therefore, the period is 2 seconds

b) and c)

We can calculate this, using the general formula of period which is:

T = 2π√L/g

where g is gravity, and in this case the free fall acceleration.

L is the length of the pendulum and T the period.

As we calculated in part a) the period is 2000 s, so solving for g we have:

T / 2π = √L/g

T² / 4π² = L/g

g (T²/4π²) = L

g = 4π²L / T²

This expression must be used to calculate g for Cambridge and Tokyo.

For Cambridge:

g = 4π² * 0.9942 / (2) ²

g = 39.249 / 4

9 = 9.8122 m/s²

For Tokyo:

g = 4π² * 0.9927 / 4

g = 9.7976 m/s²