Question

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.25 m and finds that it makes 419 complete oscillations in 1230 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 1

There's a mistake in this solution. The linear frequency should be 419/1230 = 0.34065, not 419/60 = 6.98.

Answer 2

Answer:

$g=2406.8m/s^{2}$

Explanation:

firstly we have to find the frequency and then the angular frequency.

The frequency will be:

   $f=\frac{419}{60} \\f=6.98Hz$

The angular frequency is:

ω$=2\pi f$

ω$=2*\frac{22}{7} *6.98$

ω$=43.88rad /sec$

Now we can apply simple pendulum relationship

ω$=\sqrt{\frac{g}{l} }$

make g the subject of the formula

$g=w^{2} l\\g=43.88^{2} *1.25\\g=2406.8m/s^{2}$