What is the length of a simple pendulum, which ticks seconds?
the time period of a simple pendulum is given by,
$T=2 \pi \sqrt{\frac{L}{g}}$
From this relation one gets,
$L=\frac{g T^{2}}{4 \pi^{2}}$
The time period of a simple pendulum, which ticks seconds, is $2\, s$. Therefore, for $g=9.8 \,m s ^{-2}$ and $T=2 s , L$ is
$=\frac{9.8\left( m s ^{-2}\right) \times 4\left( s ^{2}\right)}{4 \pi^{2}}$
$=1 \,m$
A uniform rod of length $2.0 \,m$ is suspended through an end and is set into oscillation with small amplitude under gravity. The time period of oscillation is approximately .... $\sec$
Time period of a simple pendulum is $T$. The angular displacement for amplitude is $\beta$. How much time the bob of pendulum will take to move from equilibrium position $O$ to $A$, making an angle $\alpha$ at the support
What effect occurs on the frequency of a pendulum if it is taken from the earth surface to deep into a mine
A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$, then the time period is given by $T = 2\pi \sqrt {\frac{l}{{g'}}} $, where $g'$ is equal to
A pendulum is suspended by a string of length $250\,cm$. The mass of the bob of the pendulum is $200\,g$. The bob is pulled aside until the string is at $60^{\circ}$ with vertical as shown in the figure. After releasing the bob. the maximum velocity attained by the bob will be________ $ms ^{-1}$. (if $g=10\,m / s ^{2}$ )