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?

Answer 1

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.