The linear displacement \( x \) of the bob of simple pendulum from its mean position varies as \( x = a \sin \left( \frac{\pi}{2} t \right) \), where \( a \) is its amplitude expressed in meter and \( t \) is in second. The length of simple pendulum is (Take \( g = \pi^2 \, \text{m/s}^2 \))
Show Hint
For simple pendulums, the angular frequency \( \omega \) is related to the length \( L \) and acceleration due to gravity \( g \) by \( \omega = \sqrt{\frac{g}{L}} \).
Step 1: Using the simple pendulum equation.
The equation for the displacement of a simple pendulum is:
\[
x = a \sin(\omega t)
\]
Where \( \omega = \sqrt{\frac{g}{L}} \) is the angular frequency and \( L \) is the length of the pendulum. Comparing this with the given equation \( x = a \sin\left( \frac{\pi}{2} t \right) \), we find:
\[
\omega = \frac{\pi}{2}
\]
Thus, equating \( \omega = \sqrt{\frac{g}{L}} \), we get:
\[
\frac{\pi}{2} = \sqrt{\frac{\pi^2}{L}}
\]
Solving for \( L \), we get \( L = 2.0 \, \text{m} \). Thus, the correct answer is (C) 2.0 m.