Question

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 0.825 m and finds that it makes 397 complete oscillations in 546 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

Answer:

17.215 m/s^2

Explanation:

L = 0.825 m

It completes 397 oscillations in 546 s.

Time period is defined as the time taken to complete one oscillation.

So, Time period, T = 546 / 397 = 1.375 s

Let g be the gravitational acceleration at that planet.

Use the formula for the time period

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

$1.375 = 2\pi \sqrt{\frac{0.825}{g}}$

$1.89 = {\frac{32.536}{g}}$

g = 17.215 m/s^2