Answer:
[tex]W = 9.081\,N[/tex]
Explanation:
The angular frequency of a simple pendulum is:
[tex]\omega = \sqrt{\frac{g}{l} }[/tex]
Where:
[tex]g[/tex] - Gravitational constant, in [tex]\frac{m}{s^{2}}[/tex].
[tex]l[/tex] - The rod length, in m.
The period of oscillation of the simple pendulum is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]
Given that the same pendulum is tested of both planets, the following relation is determined:
[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]
The gravity constant on Planet X is:
[tex]g_{2} = g_{1}\cdot \left(\frac{T_{1}}{T_{2}}\right)^{2}[/tex]
[tex]g_{2} = (9.807\,\frac{m}{s^{2}})\cdot \left(\frac{1\,s}{1.8\,s} \right)^{2}[/tex]
[tex]g_{2} = 3.027\,\frac{m}{s^{2}}[/tex]
The weight of the pendulum bob on Planet X is:
[tex]W = (3\,kg) \cdot (3.027\,\frac{m}{s^{2}} )[/tex]
[tex]W = 9.081\,N[/tex]
