Question

A simple pendulum has a bob suspended by an inextensible thread of length $1$ metre from a point $A$ of suspension. At the extreme position of oscillation, the thread is suddenly caught by a peg at a point $B$ distant (1/4) m from $A$ and the bob begins to oscillate in the new condition. The change in frequency of oscillation of the pendulum is approximately given by $\left(g=10 m / s ^{2}\right)$.

• A. $\frac{\sqrt{10}}{2}$ hertz
• B. $\frac{1}{4\sqrt{10}}$ hertz
• C. $\frac{\sqrt{10}}{3}$ hertz
• D. $\frac{1}{\sqrt{10}}$ hertz

Answer 1

Answer: (B)

$\frac{1}{4\sqrt{10}}$ hertz

Answer 2

Length of the pendulum is l = 1 m. ? Time period $T=2\pi\sqrt{\frac{l}{g}}=2\pi\sqrt{\frac{1}{10}}$ ? FrequTency $f =\frac{1}{T}=\frac{1}{2\pi}\sqrt{10}=0.5032$ Hz. Now since the pendulum thread is caught by a thread 1/4 m from the original point of suspension, so the new length of the pendulum is $l'=1-\frac{1}{4}=\frac{3}{4}$m ? Frequency $f '=\frac{1}{2\pi}\sqrt{\frac{10}{3/4}}=\frac{2}{\sqrt{3}} \frac{1}{2\pi}\sqrt{10}=0.5811$ Hz. ? Change in frequency, $\Delta f =0.5811-0.5032$ $=0.0779$ Hz.