Question

The path of oscillation of simple pendulum of length $1$ meter is $16$cm. Its maximum velocity is (g = ${\pi }^{2}m/s^2)$

• A. $2\pi cm/s$
• B. $4\pi cm/s$
• C. $8\pi cm/s$
• D. $16\pi cm/s$

Answer 1

Answer: (C)

$8\pi cm/s$

Given :- $l=1m=100cm$ and $s=16cm$

$l\times \left(2\theta \right)=s$

$⟹2\theta =\frac{s}{l}=0.16$

$⟹\theta =0.08$

Now, taking lowest point on path as reference for gravitational potential energy and applying Law of Conservation of Mechanical energy:-

Energy at extreme=Energy at mean that is bottom

$⟹mg(l-l\cos \theta )=\frac{1}{2}m{v}^2$

$⟹v=\sqrt{2gl(1-\cos \theta )}$

$⟹v=\sqrt{4gl{\sin }^2\frac{\theta }{2}}$

$⟹v=2\sqrt{gl}\sin \frac{\theta }{2}$

Since, $\theta$ is very small, $\sin \frac{\theta }{2}\approx \frac{\theta }{2}$

$⟹v=2\sqrt{gl}\frac{\theta }{2}$

$⟹v=\theta \sqrt{gl}=0.08\sqrt{{\pi }^2\times 1}$

$⟹v=0.08\pi m/s=8\pi cm/s$

Hence, answer is option-(C)