Question

A simple pendulum has a bob suspended by an inextensible thread of length 1 meter from a point A of suspension. As it is freely oscillating and at a certain stage has reached one extreme position, the thread is suddenly caught by a peg at a point B distant $\frac{1}{4}$ m from A and the bob begins to oscillate in the new condition. The change in the frequency of$(g=10m/s^2)$ oscillation of the pendulum is approximated given by

• A. $\frac{\sqrt{g}}{4\pi }(\frac{2}{\sqrt{3}}-1)$ hertz
• B. $\frac{\sqrt{g}}{2\pi }(\frac{2}{\sqrt{3}}-1)$ hertz
• C. $\frac{\sqrt{g}}{2\pi }(\frac{1}{\sqrt{3}}-1)$ hertz
• D. $\frac{\sqrt{g}}{\pi }(\frac{1}{\sqrt{3}}-1)$ hertz

Answer 1

Answer: (B)

$\frac{\sqrt{g}}{2\pi }(\frac{2}{\sqrt{3}}-1)$ hertz

$w_1=\sqrt{\frac{g}{l_1}};w_2=\sqrt{\frac{g}{l_2}}$
$l_1=1m;l_2=\frac{3}{4}m$
change in frequency =$\frac{w_2-w_1}{2\pi }$
=$\frac{\sqrt{\frac{4g}{3}}-\sqrt{g}}{2\pi }=\frac{\sqrt{g}}{2\pi }(\frac{2}{\sqrt{3}}-1)$ hertz