The time period of a simple pendulum is $2 \, sec$. If its length is increased $4$ times,then its period becomes ..... $\sec$.

Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$ from their mean positions,respectively. If the angular accelerations of both pendulums are same,then which expression is correct?

$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

The time period of a simple pendulum is $T$. If its length is increased by $21\%$,what is the percentage increase in its time period?

$A$ simple pendulum of length $1 \ m$ and having a bob of mass $100 \ g$ is suspended in a car,moving on a circular track of radius $100 \ m$ with uniform speed $10 \ m/s$. If the pendulum makes small oscillations in a radial direction about its equilibrium position,then its time period is given by $T = 2\pi / \alpha^{1/4}$. The value of $\alpha$ is (Take $g = 10 \ m/s^2$)

The periodic time of a simple pendulum of length $1\, m$ and amplitude $2\, cm$ is $5\, s$. If the amplitude is made $4\, cm$,its periodic time in seconds will be