Question

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds. A pendulum with a length of 220.5 cm has a period of 3 s.

1) Find an equation connecting l and T

2) Find the length of a pendulum which has a period of 5s.

3) What is the period of a pendulum which has a length of 0.98m?

Answer 1

1)
`L \propto T^2`

Using the condition given:

`2.205 m = K (3)^2`

`K = 0.245 \approx (g)/(4\pi^2)`

So then if we want to create an equation we need to do this:

`L = K T^2`

With K a constant. For this case the period of a pendulumn is given by this general expression:

`T = 2\pi \sqrt{(L)/(g)}`

Where L is the length in m and g the gravity
`g = 9.8 (m)/(s^2)`.

2)
`T = 2\pi \sqrt{(L)/(g)}`

If we square both sides of the equation we got:

`T^2 = 4 \pi^2 (L)/(g)`

And solving for L we got:

`L = (g T^2)/(4 \pi^2)`

Replacing we got:

`L =(9.8 (m)/(s^2) (5s)^2)/(4 \pi^2) = 6.206m`

3)
`T = 2\pi \sqrt{(0.98m)/(9.8(m)/(s^2))}= 1.987 s`

Explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

`L \propto T^2`

Using the condition given:

`2.205 m = K (3)^2`

`K = 0.245 \approx (g)/(4\pi^2)`

So then if we want to create an equation we need to do this:

`L = K T^2`

With K a constant. For this case the period of a pendulumn is given by this general expression:

`T = 2\pi \sqrt{(L)/(g)}`

Where L is the length in m and g the gravity
`g = 9.8 (m)/(s^2)`.

Part 2

For this case using the function in part a we got:

`T = 2\pi \sqrt{(L)/(g)}`

If we square both sides of the equation we got:

`T^2 = 4 \pi^2 (L)/(g)`

And solving for L we got:

`L = (g T^2)/(4 \pi^2)`

Replacing we got:

`L =(9.8 (m)/(s^2) (5s)^2)/(4 \pi^2) = 6.206m`

Part 3

For this case using the function in part a we got:

`T = 2\pi \sqrt{(L)/(g)}`

Replacing we got:

`T = 2\pi \sqrt{(0.98m)/(9.8(m)/(s^2))}= 1.987 s`