A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _____ s. [take \( g = \pi^2 \, m/s^2 \)]
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _____ s. [take \( g = \pi^2 \, m/s^2 \)]
Correct Answer: 8
Solution and Explanation
The time period of a pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}}, \] where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
For the given scenario, the effective acceleration due to gravity is:
\[ g' = \frac{g}{4}. \]
Substituting this into the formula for the time period:
\[ T = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4L}{g}}. \]
Given that \( L = 4 \, m \), the formula becomes:
\[ T = 2\pi \sqrt{\frac{4 \times 4}{g}}. \]
Simplifying further:
\[ T = 2\pi \sqrt{\frac{16}{g}}. \]
Given \( g = \pi^2 \, m/s^2 \), substitute into the equation:
\[ T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \times \frac{4}{\pi}. \]
Simplify:
\[ T = 2 \times 4 = 8 \, s. \]
Final Answer:
\[ T = 8 \, s. \]
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