
Total length of a pendulum remains same, so extreme point $D$ lies on the line $A B$, as shown below.
This can be proved by applying energy conservation between extreme positions $A$ and $D$ (its given friction is abscent),
$K_A+U_A=K_B+U_B=K_D+U_D$
$\Rightarrow 0+U_A=0+U_B=0+U_D$
$\Rightarrow U_A=U_B=U_D \Rightarrow h_A=h_B=h_D$
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .
| Column $I$ | Column $II$ |
| $(A)$ Potential energy of a simple pendulum (y axis) as a function of displacement ( $\mathrm{x}$ axis) | $Image$ |
| $(B)$ Displacement (y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive $\mathrm{x}$-direction | $Image$ |
| $(C)$ Range of a projectile (y axis) as a function of its velocity ( $\mathrm{x}$ axis) when projected at a fixed angle | $Image$ |
| $(D)$ The square of the time period (y axis) of a simple pendulum as a function of its length ( $\mathrm{x}$ axis) | $Image$ |
