Question

on another planet a simple pendulum has a length of 90.0 cm oscillates with a period of 1.7 seconds.

1. compute the planet’s accerleration due to gravity and comment on the planet’s gravity as compared to Earth’s gravity.

2. on this same planet. what would be the period of oscillation if the pendulum had a length of exactly 2.00 meters?

Answer 1

Time period of oscillation of a simple pendulum is T = 2 where L (m) is length of pendulum and 'g' is acceleration due to gravity in m/s^2 . Given length = 90*10^(-2)m , time period = 1.7 s rearranging above equation we get,

1) g = L* (2 / T)^2 = 90* 10^(-2) * (2* 3.14 /1.7)^2 = 12.29m/s^2. The acceleration due to gravity on earth is 9.8 m/s^2. The value of g on this planet is more than that of earth. It is nearly 1.25 times the acceleration due to gravity on earth.

2)when length becomes L=2m g = 12.29 m/s^2 using the formula T= 2

T = 2* =   2.53 s . From the formula time period is directly proportional to square root of length of pendulum and inversely proportional to square root of acceleration due to gravity. As length increases the time period also increases.