Question

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441 complete oscillations in 1090 s. the amplitude of the oscillations is very small compared to the pendulum's length. what is the gravitational acceleration on the surface of this planet? answer in units of m/s 2 .

Answer 1

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$