Physics: 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

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A simple pendulum is made my attaching a rod

Ответ:

serenity6282

24.01.2022

Physics

$W = 9.081\,N$

Explanation:

The angular frequency of a simple pendulum is:

$\omega = \sqrt{\frac{g}{l} }$

Where:

- Gravitational constant, in $\frac{m}{s^{2}}$ .

- The rod length, in m.

The period of oscillation of the simple pendulum is:

$T = 2\pi\cdot \sqrt{\frac{l}{g} }$

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:

$g_{2} = g_{1}\cdot \left(\frac{T_{1}}{T_{2}}\right)^{2}$

$g_{2} = (9.807\,\frac{m}{s^{2}})\cdot \left(\frac{1\,s}{1.8\,s} \right)^{2}$

$g_{2} = 3.027\,\frac{m}{s^{2}}$

The weight of the pendulum bob on Planet X is:

$W = (3\,kg) \cdot (3.027\,\frac{m}{s^{2}} )$

$W = 9.081\,N$

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