An oscillating pendulum suspended from the ro | 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 $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.For a lift at rest, $g_{eff} = g$, so $T_1 = 2\pi \sqrt{\frac

Vedclass · Accepted Answer

$T_1 = \frac{\sqrt{2}T_2T_3}{\sqrt{T_2^2 + T_3^2}}$

Vedclass · Answer

(B) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.For a lift at rest, $g_{eff} = g$, so $T_1 = 2\pi \sqrt{\frac{l}{g}}$.When the lift moves up with acceleration $a$, $g_{eff} = g + a$, so $T_2 = 2\pi \sqrt{\frac{l}{g+a}}$.When the lift moves down with acceleration $a$, $g_{eff} = g - a$, so $T_3 = 2\pi \sqrt{\frac{l}{g-a}}$.Squaring the equations, we get $T_2^2 = 4\pi^2 \frac{l}{g+a}$ and $T_3^2 = 4\pi^2 \frac{l}{g-a}$.Taking the reciprocals: $\frac{1}{T_2^2} = \frac{g+a}{4\pi^2 l}$ and $\frac{1}{T_3^2} = \frac{g-a}{4\pi^2 l}$.Adding these: $\frac{1}{T_2^2} + \frac{1}{T_3^2} = \frac{g+a+g-a}{4\pi^2 l} = \frac{2g}{4\pi^2 l} = \frac{2}{T_1^2}$.Thus, $\frac{T_3^2 + T_2^2}{T_2^2 T_3^2} = \frac{2}{T_1^2}$, which implies $T_1^2 = \frac{2T_2^2 T_3^2}{T_2^2 + T_3^2}$.Taking the square root, we get $T_1 = \frac{\sqrt{2}T_2T_3}{\sqrt{T_2^2 + T_3^2}}$.

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$. When the lift moves up with acceleration $a$, its time period is $T_2$. When the lift moves down with acceleration $a$, its time period is $T_3$. The relation between $T_1$, $T_2$, and $T_3$ is:

Similar Questions

$A$ simple pendulum oscillating in air has period $T.$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$ of the material of the bob. If the bob is inside the liquid all the time,its period of oscillation in this liquid is:

If the length of a simple pendulum is increased by $300\%$,then the time period will be increased by ..... $\%$

The time period of a simple pendulum is $T$. If its length is increased by $21\%$,what is the percentage increase in its time period?

$A$ simple pendulum has a time period $T$. The bob is given a negative charge and the surface below it is given a positive charge. The new time period will be

If a body dropped freely from a height of $20 \,m$ reaches the surface of a planet with a velocity of $31.4 \,ms^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is (in $\,m$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module