A second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket:

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1. **Formula:** The time period $T$ of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$, where $L$ is the length and $g_{eff}$ is the effective acceleration due to gravity .
2. **Analysis:** To decrease the period $T$, the effective gravity $g_{eff}$ must increase (since $T \propto \frac{1}{\sqrt{g_{eff}}}$).
3. **Case 1 (Moving up with acceleration $a$):** When the rocket accelerates upwards, the pseudo-force acts downwards, adding to gravity. Thus, $g_{eff} = g + a$. Since $g_{eff} > g$, the period $T$ decreases.
4. **Case 2 (Moving down with acceleration $a$):** Pseudo-force acts upwards. $g_{eff} = g - a$. $T$ increases.
5. **Case 3 (Uniform velocity):** Acceleration $a = 0$. $g_{eff} = g$. Period remains unchanged.
6. **Case 4 (Geostationary orbit):** Inside a satellite orbiting the earth, the effective gravity is zero ($g_{eff} \approx 0$). The pendulum would not oscillate ($T 	o \infty$).

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