Question

A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the period of small-angle oscillations is 1.0 second. It is also observed that on Planet X this same pendulum has a period of 1.8 seconds. How much does the pendulum bob weigh on Planet X

Answer 1

`W = 9.081\,N`

Step-by-step explanation:

The angular frequency of a simple pendulum is:

`\omega = \sqrt{(g)/(l) }`

Where:

`g` - Gravitational constant, in
`(m)/(s^(2))`.

`l` - The rod length, in m.

The period of oscillation of the simple pendulum is:

`T = 2\pi\cdot \sqrt{(l)/(g) }`

Given that the same pendulum is tested of both planets, the following relation is determined:

`T_(1)^(2)\cdot g_(1) = T_(2)^(2)\cdot g_(2)`

The gravity constant on Planet X is:

`g_(2) = g_(1)\cdot \left((T_(1))/(T_(2))\right)^(2)`

`g_(2) = (9.807\,(m)/(s^(2)))\cdot \left((1\,s)/(1.8\,s) \right)^(2)`

`g_(2) = 3.027\,(m)/(s^(2))`

The weight of the pendulum bob on Planet X is:

`W = (3\,kg) \cdot (3.027\,(m)/(s^(2)) )`

`W = 9.081\,N`