The ratio of the frequencies of two simple pe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The frequency $f$ of a simple pendulum is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$,where $g$ is the acceleration due to gravit

Vedclass · Accepted Answer

$9: 16$

Vedclass · Answer

(C) The frequency $f$ of a simple pendulum is given by the formula $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$,where $g$ is the acceleration due to gravity and $L$ is the length of the pendulum.Since the pendulums are at the same place,$g$ is constant.Therefore,$f \propto \frac{1}{\sqrt{L}}$,which implies $f^2 \propto \frac{1}{L}$ or $L \propto \frac{1}{f^2}$.Given the ratio of frequencies $f_1 : f_2 = 4 : 3$,we have $\frac{f_1}{f_2} = \frac{4}{3}$.The ratio of their lengths is $\frac{L_1}{L_2} = \frac{f_2^2}{f_1^2} = \left( \frac{f_2}{f_1} \right)^2$.Substituting the values,$\frac{L_1}{L_2} = \left( \frac{3}{4} \right)^2 = \frac{9}{16}$.Thus,the ratio of their lengths is $9: 16$.

The ratio of the frequencies of two simple pendulums is $4: 3$ at the same place. The ratio of their respective lengths is

Similar Questions

$A$ $metre$ scale is suspended vertically from a horizontal axis passing through one end of it. Its time period would be ....... $\sec$

The angular velocity and the amplitude of a simple pendulum are $\omega$ and $A$ respectively. At a displacement $x$ from the mean position,its kinetic energy is $T$ and potential energy is $V$. Then the ratio $\frac{V}{T}$ is

$A$ simple pendulum executing $S.H.M.$ is falling freely along with its support. Then:

Is the oscillation of a simple pendulum at the centre of the earth possible?

Define a simple pendulum and the length of the pendulum.

Vedclass Test Series

Exam Paper Generator

Online Exam Module