A pendulum that has a period of 3.00s and that is located where the acceleration due to gravity is 9.79 m/s^2 is moved to a location where the acceleration due to gravity is 9.82 m/s^2. What is its new period, in s?
The period of a pendulum can be calculated using the formula:
T = 2π√(L / g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In this case, the period T1 is given as 3.00s and the acceleration due to gravity g1 is 9.79 m/s^2. Let's assume the length L is constant.
Using the formula, we can write:
T1 = 2π√(L / g1)
Rearranging the equation to solve for L:
L = (T1^2 * g1) / (4π^2)
Now we can use the same formula to find the new period T2 with the new acceleration due to gravity g2:
T2 = 2π√(L / g2)
Substituting the value of L from the previous equation:
T2 = 2π√(((T1^2 * g1) / (4π^2)) / g2)
Simplifying:
T2 = π√((T1^2 * g1) / (π^2 * g2))
T2 = (T1 / π) * √(g1 / g2)
Plugging in the values, T1 = 3.00s, g1 = 9.79 m/s^2, and g2 = 9.82 m/s^2:
T2 = (3.00 / π) * √(9.79 / 9.82)
T2 ≈ 1.91s
Therefore, the new period of the pendulum, when located where the acceleration due to gravity is 9.82 m/s^2, is approximately 1.91 seconds.