The time period of a simple pendulum of length $L$ is $T_1$. The time period of a uniform rod of the same length $L$ suspended from one end and oscillating in a vertical plane is $T_2$. The amplitude of oscillation is small in both cases. Then the ratio $\frac{T_1}{T_2}$ is:

If the length of a pendulum is made $9$ times and the mass of the bob is made $4$ times,then the value of the time period becomes:

$A$ pendulum is oscillating at a frequency of $8 \,Hz$. Suddenly, the string of the pendulum is clamped at its midpoint. What is the new frequency of oscillations (in $\,Hz$)?

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length,then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is

In an experiment for determining the gravitational acceleration $g$ of a place with the help of a simple pendulum,the measured time period square $(T^2)$ is plotted against the string length $(L)$ of the pendulum as shown in the figure. What is the value of $g$ at the place in $m/s^2$?

$A$ 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 $g = 10 \ m/s^2$)