A simple pendulum is executing simple harmoni | vedclass

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

Question

(a) If initial length ${l_1} = 100$ then ${l_2} = 121$

By using $T = 2\pi \sqrt {\frac{l}{g}} $

$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\f

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(a) If initial length ${l_1} = 100$ then ${l_2} = 121$

By using $T = 2\pi \sqrt {\frac{l}{g}} $

$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{l_1}}}{{{l_2}}}} $

Hence, $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{100}}{{121}}} $

$\Rightarrow {T_2} = 1.1\,{T_1}$

$\%$ increase = $\frac{{{T_2} - {T_1}}}{{{T_1}}} 	imes 100 = 10\,\% $

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

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

Which of the following statements is not true ? In the case of a simple pendulum for small amplitudes the period of oscillation is

A solid cylinder of density $\rho_0$, cross-section area $A$ and length $l$ floats in a liquid of density $\rho\left( >\rho_0\right)$ with its axis vertical, as shown. If it is slightly displaced downward and released, the time period will be .......

What happens to the time period of a simple pendulum when it is taken to moon's surface from earth's surface ?