The total energy of a particle executing simp | 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 kinetic energy $(K)$ and potential energy $(U)$.$

Vedclass · Accepted Answer

independent of $x$

Vedclass · Answer

(C) The total energy $(E)$ of a particle executing simple harmonic motion is given by the sum of its kinetic energy $(K)$ and potential energy $(U)$.$E = K + U = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.Since $m$,$\omega$,and $A$ are constants for a given simple harmonic motion,the total energy $E$ remains constant throughout the motion.Therefore,the total energy is independent of the displacement $x$ from the equilibrium position.Thus,option $(c)$ is correct.

The total energy of a particle executing simple harmonic motion is

Similar Questions

Identify the odd one out.

What is the product formed when ethyl alcohol reacts with chlorine in the presence of $NaOH$?

If the circle $x^2+y^2+8x-4y+c=0$ touches the circle $x^2+y^2+2x+4y-11=0$ externally and cuts the circle $x^2+y^2-6x+8y+k=0$ orthogonally,then $k=$

If the lines $x^2+2xy-35y^2-4x+44y-12=0$ and $5x+\lambda y-8=0$ are concurrent,then $\lambda$ equals

The oxoacid of sulphur which contains two sulphur atoms in different oxidation states is

Vedclass Test Series

Exam Paper Generator

Online Exam Module