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

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

Answer:

$L = 1.24\ m$

Explanation:

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

$T=2\pi \sqrt(\frac{L}{g})$,

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

Solving For the length we get:

$L=g( \frac{T}{2\pi})^{2}$.

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:

$L=g(\frac{T}{2\pi})^{2}\\\\L=49(\dfrac{1}{2\pi})^{2}\\\\\\\\L=1.24\ m$