In Tokyo, Japan, a seconds pendulum is 0.9914 m long. What is the free-fall acceleration in Tokyo?1. 9.81 m/s^2 2. 9.81441 m/s^2 3.9.78473 m/s^2 4. 9.78366 m/s^2 5. None of these 6. 9.81335 m/s^2

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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 Tokyo, Japan is 0.9914 m, it gives:

2 = 2*pi*sqrt(L/g)

=> sqrt(L/g) = 1/pi

=> L/g = 1/pi^2

=> g = L*pi^2

=> g = 0.9914*pi^2

=> g = 9.78473 m/s^2

The correct answer is option 3, 9.78473 m/s^2

senioreeto's profile pic

senioreeto | Student, Undergraduate | (Level 1) Valedictorian

Posted on

the answer is 9.78473 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=0.9914m.

T=2 sec.

solving the above equation for'g',we get the value of 'g' as 9.78473 m/s2 which is option (3).

We’ve answered 318,961 questions. We can answer yours, too.

Ask a question