A mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pe

Question

A mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the earth, it is found that $t=2T$. The entire setup is taken on the surface of another planet whose mass is half of that of the Earth and whose radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$. Then we can say:

Accepted Answer

1. **Formulas:** 
   - Time of free fall from height $h$: $t = \sqrt{\frac{2h}{g}}$ (derived from kinematic equation $h = \frac{1}{2}gt^2$). [Chapter 2, Example 2.4]
   - Time period of a simple pendulum: $T = 2\pi\sqrt{\frac{l}{g}}$. [Chapter 1, Example 1.5]
   - Acceleration due to gravity: $g = \frac{GM}{R^2}$. [Chapter 7, Eq. 7.9]

2. **Dependency on $g$:** 
   Both $t$ and $T$ are inversely proportional to $\sqrt{g}$.
   $$ t \propto \frac{1}{\sqrt{g}} \quad 	ext{and} \quad T \propto \frac{1}{\sqrt{g}} $$

3. **Ratio Analysis:** 
   The ratio of the time of fall to the time period is independent of $g$:
   $$ \frac{t}{T} = \frac{\sqrt{2h/g}}{2\pi\sqrt{l/g}} = \frac{\sqrt{2h}}{2\pi\sqrt{l}} = 	ext{constant} $$
   Since $h$ and $l$ (length of pendulum) remain constant in the setup, this ratio remains constant regardless of the value of $g$.

4. **Conclusion:** 
   On Earth, $\frac{t}{T} = 2$. On the other planet, despite the change in mass ($M' = M/2$) and resulting gravity ($g' = g/2$), the ratio must remain the same.
   $$ \frac{t'}{T'} = \frac{t}{T} = 2 \implies t' = 2T' $$

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