A certain pendulum oscillates back and forth every 3.0 s. What would the new period be, in s, if the pendulum were one-half of its original length?
The period of a pendulum is given by 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.
Given that the original period is 3.0 s, let's assume the original length of the pendulum is L.
Using the formula above, we can write:
3.0 = 2π√(L/g).
Now, let's consider the new length of the pendulum, which is half of the original length. Therefore, the new length is L/2.
Using the same formula, we can write the new period as:
T' = 2π√[(L/2)/g].
To find the new period T', we need to solve for L/2:
3.0 = 2π√[(L/2)/g].
To simplify the equation, let's square both sides:
(3.0)^2 = (2π)^2[(L/2)/g].
9.0 = 4π^2(L/2g).
Now, let's solve for L/2. We have:
L/2 = 9.0/(4π^2/g).
L/2 = 9.0g/(4π^2).
Multiplying both sides by 2:
L = 18.0g/(4π^2).
Now, let's substitute the new length into the formula for the new period. We have:
T' = 2π√[(L/2)/g].
T' = 2π√[(18.0g/(4π^2)/g].
T' = 2π√[18.0/(4π^2)].
T' = 2π√(9/(2π)^2).
T' = 2π(3/(2π)).
T' = 6.0 s.
Therefore, the new period, if the pendulum were half of its original length, would be 6.0 seconds.