What length pendulum will produce a period of 1.0s?
t = 2p1*sqrt(l/g),
t^2 = 4 * (pi)^2 * l/g.
t in seconds.
l in meters or cm.
g = acceleration due to gravity:
9.8 m/s^2 or 980 cm/s^2.
NOTE: l and g should both be in m/s^2
or cm/s^2.
t^2 = 4 * 9.87 * l/9.8 = 1^2,
39.48l / 9.8 = 1,
4.03l = 1,
l = 1 / 4.03 = 0.248 m = 24.8 cm.
To determine the length of a pendulum that will produce a period of 1.0 second, we need to use the pendulum's mathematical formula. The formula for the period of a simple pendulum is given by:
T = 2π * √(L / g)
Where:
T is the period of the pendulum,
L is the length of the pendulum, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, we are given that the period T is 1.0 second. So, we can rearrange the formula and solve for L:
1.0 = 2π * √(L / 9.8)
First, divide both sides of the equation by 2π:
1.0 / (2π) = √(L / 9.8)
Next, square both sides of the equation to eliminate the square root:
(1.0 / (2π))^2 = L / 9.8
Simplifying further:
1.0 / (2π)^2 = L / 9.8
Finally, multiply both sides of the equation by 9.8:
1.0 / (2π)^2 * 9.8 = L
Evaluating the expression on the left side:
(1.0 / (2π)^2) * 9.8 ≈ 0.249 m
Therefore, a pendulum with a length of approximately 0.249 meters (or 24.9 centimeters) will produce a period of 1.0 second.