A particle of mass m is executing simple harm | vedclass

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

Question

(C) The total energy $(E)$ of a particle executing simple harmonic motion is given by the sum of its potential energy and kinetic energy.The formula f

Vedclass · Accepted Answer

$\frac{2 \pi^{2} m A^{2}}{T^{2}}$

Vedclass · Answer

(C) The total energy $(E)$ of a particle executing simple harmonic motion is given by the sum of its potential energy and kinetic energy.The formula for total energy is $E = \frac{1}{2} k A^{2}$,where $k$ is the force constant.We know that the angular frequency $\omega = \frac{2 \pi}{T}$.Also,the force constant $k = m \omega^{2}$.Substituting the value of $\omega$ into the expression for $k$,we get $k = m \left( \frac{2 \pi}{T} \right)^{2} = \frac{4 \pi^{2} m}{T^{2}}$.Now,substitute $k$ into the total energy formula: $E = \frac{1}{2} \left( \frac{4 \pi^{2} m}{T^{2}} \right) A^{2}$.Simplifying this,we get $E = \frac{2 \pi^{2} m A^{2}}{T^{2}}$.

$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is

Similar Questions

$A$ particle is executing simple harmonic motion with amplitude $A$. The ratio of the kinetic energies of the particle when it is at displacements of $\frac{A}{4}$ and $\frac{A}{2}$ from the mean position is

$A$ particle of mass $4 \,kg$ is executing $SHM$. Its displacement is given by the equation $y=8 \cos [100 t+\pi / 4] \,cm$. Its maximum kinetic energy is (in $\,J$)

$A$ block is executing $SHM.$ Let the time period of variation of velocity be $T_1$ and time period of variation of kinetic energy be $T_2.$ The relation between $T_1$ and $T_2$ is

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .

$A$ particle is vibrating in $S.H.M.$ with an amplitude of $4 \,cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

Vedclass Test Series

Exam Paper Generator

Online Exam Module