The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $\frac{g}{3}$ is
$2\pi \sqrt {\frac{{3L}}{g}} $
$\pi \sqrt {\left( {\frac{{3L}}{g}} \right)} $
$2\pi \sqrt {\left( {\frac{{3L}}{{2g}}} \right)} $
$2\pi \sqrt {\frac{{2L}}{{3g}}} $
On a planet a freely falling body takes $2 \,sec$ when it is dropped from a height of $8 \,m$, the time period of simple pendulum of length $1\, m$ on that planet is ..... $\sec$
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical, the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums ?
''Motion of simple pendulum from mean position for small displacement is a simple harmonic motion'' - Explain this statement.
A simple pendulum of frequency $f$ has a metal bob. If bob is charged negatively and is allowed to oscillate with large positive charged plate under it, frequency will be
A simple pendulum executing $S.H.M.$ is falling freely along with the support. Then