The angular velocity and the amplitude of a s | vedclass

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

Question

(D) For a simple harmonic oscillator,the kinetic energy $T$ at displacement $X$ is given by $T = \frac{1}{2} m \omega^2 (a^2 - X^2)$.The potential ene

Vedclass · Accepted Answer

$(a^2 - X^2) / X^2$

Vedclass · Answer

(D) For a simple harmonic oscillator,the kinetic energy $T$ at displacement $X$ is given by $T = \frac{1}{2} m \omega^2 (a^2 - X^2)$.The potential energy $V$ at displacement $X$ is given by $V = \frac{1}{2} m \omega^2 X^2$.To find the ratio of kinetic energy $T$ to potential energy $V$,we divide the two expressions:$\frac{T}{V} = \frac{\frac{1}{2} m \omega^2 (a^2 - X^2)}{\frac{1}{2} m \omega^2 X^2}$.Canceling the common terms $\frac{1}{2} m \omega^2$,we get:$\frac{T}{V} = \frac{a^2 - X^2}{X^2}$.

The angular velocity and the amplitude of a simple pendulum are $\omega$ and $a$ respectively. At a displacement $X$ from the mean position,if its kinetic energy is $T$ and potential energy is $V$,then the ratio of $T$ to $V$ is:

Similar Questions

$A$ simple pendulum is hanging from the ceiling of a lift. When the lift is at rest,the time period of the pendulum is $T$. If the resultant acceleration becomes $g/4$,then the new time period of the pendulum is: (in $, T$)

$A$ simple pendulum of length $L$ has mass $M$ and it oscillates freely with amplitude $A$. At the extreme position,its potential energy is:

If a simple pendulum is taken to a place where $g$ decreases by $2\%$,then the time period:

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 $\,m$)

The length of a seconds pendulum is .... $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module