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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Oscillations

The time period of oscillation in sec of a simple pendulum of length l suspended inside a rocket moving vertically with an acceleration '2a' just above the earth is:

$\begin{array}{1 1}(a)\;0\\(b)\;2 * \frac{\pi}{\frac{\sqrt{(2a+g)}}{I}} \\( c)\; 2 * \frac{\pi}{\frac{\sqrt{(g-a)}}{I}} \\(d)\;\frac{\sqrt{(2a+g)}}{I} \end{array}$

1 Answer

The net accleration felt by mass m of the pendulum is $a+g$ .
Hence b is the correct answer.
answered Dec 11, 2013 by meena.p
 

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