Question

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

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

Answer 1

The correct option is B $4\pi$ $cm/s$

Since path length of $1$ complete oscillation is $16cm$

Therefore$,$ length $AOB$ will be $8cm$

Now$,$ angle $\theta =$ $\frac{arc}{Radius}$

$\theta =$ $\frac{8cm}{100cm}=0.08$

So$,$ $\frac{\theta }{2}=$ $\frac{0.08}{2}=0.04$ $radians$

So$,$ h$=$ $\left(100cm\right)$ $\cos \frac{\theta }{2}=$ $\left(100cm\right)\times \cos \left(0.04\right)$

$h=100cm\times 0.9992$

$h=99.92cm$

Therefore$,$ $△x=100cm-h=100-99.92=0.08cm$

Now all of the potential energy at $A$ will be converted to kinetic energy at $O$

$P.E=K.E$

So$,$ $mg△x=\frac{1}{2}m{v}^2$

Now$,$ given $g=x^2=\frac{1}{2}{V}^2$

$=100x^{2}cm/s^2$

$[$ since$,$ $1m=100cm]$

So$,$ $g△x=\frac{1}{2}{V}^2$

$V^2=2g△x$

$V^2=2\times 100{\pi }^{2}cm/s^2\times 0.08cm$

$V^2=\left(2\times 100\pi \times \frac{8}{100}\right)\frac{c{m}^2}{s^2}$

$V^2=16{\pi }^{2}c{m}^2/s^2$

$V=4\pi cm/s$

Hence$,$

Option $\left(B\right)$ is correct