A simple pendulum has a time period T_1. The | vedclass

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

Question

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,the point of suspension is at rest,so $g_{eff} = g

Vedclass · Accepted Answer

$\frac{6}{5}$

Vedclass · Answer

(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,the point of suspension is at rest,so $g_{eff} = g = 10 \ m/s^2$. Thus,$T_1 = 2\pi \sqrt{\frac{L}{g}}$.When the point of suspension moves upward with an acceleration $a$,the effective acceleration is $g_{eff} = g + a$.Given the displacement $y = kt^2$,the acceleration $a$ is the second derivative of displacement with respect to time: $a = \frac{d^2y}{dt^2} = \frac{d^2}{dt^2}(kt^2) = 2k$.Given $k = 1 \ m/s^2$,we have $a = 2(1) = 2 \ m/s^2$.Therefore,$g_{eff} = g + a = 10 + 2 = 12 \ m/s^2$.The new time period is $T_2 = 2\pi \sqrt{\frac{L}{g_{eff}}} = 2\pi \sqrt{\frac{L}{12}}$.Now,the ratio $\frac{T_1^2}{T_2^2} = \frac{4\pi^2 (L/g)}{4\pi^2 (L/g_{eff})} = \frac{g_{eff}}{g} = \frac{12}{10} = \frac{6}{5}$.

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1 \ m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be (take $g = 10 \ m/s^2$):

Similar Questions

$A$ child swinging on a swing in a sitting position stands up. What will happen to the time period of the swing?

$Assertion:$ The time period of a simple pendulum on a satellite orbiting the Earth is infinity.
$Reason:$ The time period of a simple pendulum is inversely proportional to $\sqrt{g}$.

If a simple pendulum is released from the given position,find the velocity of the bob when it reaches the lowest position. (Take $g = 10 \ m/s^2$) (in $m/s$)

$A$ simple pendulum is taken to a place where its distance from the Earth's surface is equal to the radius of the Earth. Calculate the time period of small oscillations if the length of the string is $4.0 \ m$. (Take $g = \pi^2 \ m/s^2$ at the surface of the Earth.) (in $s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module