The time period of a simple pendulum of lengt | vedclass

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

Question

(C) The effective acceleration $g_{eff}$ of a simple pendulum in an elevator descending with an acceleration $a = \frac{g}{3}$ is given by $g_{eff} =

Vedclass · Accepted Answer

$2\pi \sqrt{\frac{3L}{2g}}$

Vedclass · Answer

(C) The effective acceleration $g_{eff}$ of a simple pendulum in an elevator descending with an acceleration $a = \frac{g}{3}$ is given by $g_{eff} = g - a$.Substituting the given values,we get $g_{eff} = g - \frac{g}{3} = \frac{2g}{3}$.The formula for the time period $T$ of a simple pendulum is $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Substituting $g_{eff} = \frac{2g}{3}$ into the formula,we get $T = 2\pi \sqrt{\frac{L}{2g/3}} = 2\pi \sqrt{\frac{3L}{2g}}$.

The time period of a simple pendulum of length $L$ as measured in an elevator descending with acceleration $\frac{g}{3}$ is

Similar Questions

$A$ pendulum of length $l = 1\,m$ is released from $\theta_0 = 60^\circ$. The rate of change of speed of the bob at $\theta = 30^\circ$ is ........ $m/s^2$ $(g = 10\,m/s^2)$.

$A$ pendulum is oscillating at a frequency of $8 \,Hz$. Suddenly, the string of the pendulum is clamped at its midpoint. What is the new frequency of oscillations (in $\,Hz$)?

The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).

If the length of a second's pendulum is decreased by $2\%$,how many seconds will it lose per day?

In a simple pendulum,the period of oscillation $T$ is related to the length of the pendulum $l$ as

Vedclass Test Series

Exam Paper Generator

Online Exam Module