Question

This figure shows the angular displacement of a pendulum on a planet with five times the earth's gravity. How long is the pendulum's string?

1.0 m

3.2 m

0.80 m

4.0 m

Answer 1

Step-by-step explanation:

The period (time per cycle) is 1.0 s. The gravity is 5g or 49 m/s². Therefore:

T = 2π √(L / g)

1.0 s = 2π √(L / 49 m/s²)

L = 1.2 m

Answer 2

`L = 1.24\ m`

Step-by-step explanation:

It is known that the oscillation period of a pendulum can be described as

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

where T is the oscillation period, L is the length of the pendulum and, g is the gravity.

Solving For the length we get:

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

We know that g equals 5 times earth's gravity,

`g=5*9.8=49\ m/s^(2)`,

and from the angular displacement graphics, it can be seen that the period is

`T= 1\ s`.

Now, we can easily compute the length of the pendulum:

`L=g((T)/(2\pi))^(2)\\\\L=49((1)/(2\pi))^(2)\\\\\\\\L=1.24\ m\\`