Question

A simple pendulum consists of a bob of mass $m$ and a light string of length $l$ as shown in the figure. Another identical ball moving with a small velocity $v_0$ collides with the pendulum's bob and sticks to it. For this new pendulum of mass $2m$, which of the following statement(s) is/are correct ?

• A. Time period of the pendulum is $2\pi \sqrt{\frac{l}{g}}$.
• B. The equation of motion for this pendulum is $\theta =\frac{v_0}{2\sqrt{gL}}\sin \left[\sqrt{\frac{g}{l}}t\right]$.
• C. The equation of motion for this pendulum is $\theta =\frac{v_0}{4\sqrt{gl}}\cos \left[\sqrt{\frac{g}{l}}t\right]$.
• D. Time period of the pendulum is $2\pi \sqrt{\frac{2l}{g}}$.

Answer 1

Answer: (A), (B)

The equation of motion for this pendulum is $\theta =\frac{v_0}{2\sqrt{gL}}\sin \left[\sqrt{\frac{g}{l}}t\right]$.

The time period of simple harmonic pendulum is independent of mass, so it would be same as that $T=2\pi \sqrt{\frac{l}{g}}$.
After collision, the combined mass acquires a velocity of $\frac{v_0}{2}$. as a result of this velocity, the mass $\left(2m\right)$ moves up and at an angle ${\theta }_0$ (say) with vertical, it stops, this is the extreme position of bob.


From principle of conservation of energy,
$\Delta K+\Delta U=0$
$0-\frac{2m}{2}{\left(\frac{v_0}{2}\right)}^2=-2mgl(1-\cos {\theta }_0)$
$\frac{v_0^2}{8gl}=1-\cos {\theta }_0=2{\sin }^2\frac{{\theta }_0}{2}$
$\sin \frac{{\theta }_0}{2}=\frac{v_0}{4\sqrt{gl}}$
If ${\theta }_0$ is small, $\sin \frac{{\theta }_0}{2}=\frac{{\theta }_0}{2}$
$\Rightarrow {\theta }_0=\frac{v_0}{2\sqrt{gl}}$
So, the equation of simple harmonic motion is $\theta ={\theta }_0\sin \left(\omega t\right)$
$\theta =\frac{v_0}{2\sqrt{gl}}\sin \left(\omega t\right)$
Thus, options (a) and (b) are correct.