A simple pendulum is executing simple harmoni | vedclass

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

Question

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.This implies $T \propto \sqrt{l}$.Let the initial length be $l_1 =

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.This implies $T \propto \sqrt{l}$.Let the initial length be $l_1 = 100$ units. Then the new length $l_2 = 100 + 21 = 121$ units.The ratio of the time periods is $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{121}{100}} = \frac{11}{10} = 1.1$.Thus,$T_2 = 1.1 T_1$.The percentage increase in the time period is given by $\frac{T_2 - T_1}{T_1} 	imes 100 = \frac{1.1 T_1 - T_1}{T_1} 	imes 100 = 0.1 	imes 100 = 10\%$.Therefore,the correct option is $A$.

$A$ simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$,the percentage increase in the time period of the pendulum of increased length is ..... $\%$

Similar Questions

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing $SHM$ with frequency relation $f_{A}=2 f_{B}$,then the relation is:

If the metal bob of a simple pendulum is replaced by a wooden bob, then its time period will

$A$ second's pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket

$A$ lift is descending with an acceleration of $g/3$. What will be the time period of a simple pendulum suspended from its ceiling if its time period in a stationary lift is $T$?

The length of a simple pendulum is increased by $2\%$. Its time period will

Vedclass Test Series

Exam Paper Generator

Online Exam Module