In an experiment to determine the period of a | vedclass

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

Question

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length.Taking the natural logarithm and

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length.Taking the natural logarithm and differentiating,we get $\frac{dT}{T} = \frac{1}{2} \frac{dl}{l}$.Here,the effective length $l$ is the length of the string plus the radius of the bob. Thus,the change in effective length due to the change in radius is $dl = |r_1 - r_2|$.Given the relative difference in periods $\frac{\Delta T}{T} = 5 	imes 10^{-4}$ and the length $l = 1\, m$.Substituting these values into the relation $\frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}$:$5 	imes 10^{-4} = \frac{1}{2} 	imes \frac{|r_1 - r_2|}{1}$.$|r_1 - r_2| = 2 	imes 5 	imes 10^{-4} = 10 	imes 10^{-4} = 10^{-3}\, m$.Converting to centimeters: $10^{-3}\, m = 10^{-1}\, cm = 0.1\, cm$.

Similar Questions

State the laws of a simple pendulum.

The bob of a simple pendulum of length $L$ is released from a position of small angular displacement $\theta$. Its linear displacement at time $t$ is ($g =$ acceleration due to gravity).

The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with an acceleration of $g/3$,then the time period of the pendulum is

The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module