A particle is vibrating simple harmonically w | vedclass

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

Question

(B) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$.Given that the kinetic

Vedclass · Accepted Answer

$\frac{a}{\sqrt{2}}$

Vedclass · Answer

(B) The total energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$.Given that the kinetic energy is equal to the potential energy,$K = U$.Since the total energy $E = K + U$,we have $E = 2U$ or $E = 2K$.The potential energy at displacement $y$ is given by $U = \frac{1}{2} m \omega^2 y^2$.The total energy is given by $E = \frac{1}{2} m \omega^2 a^2$.Setting $U = \frac{1}{2} E$,we get:$\frac{1}{2} m \omega^2 y^2 = \frac{1}{2} (\frac{1}{2} m \omega^2 a^2)$$y^2 = \frac{a^2}{2}$$y = \frac{a}{\sqrt{2}}$

$A$ particle is vibrating simple harmonically with an amplitude $a$. The displacement of the particle when its energy is half kinetic and half potential is

Similar Questions

$A$ particle is executing $SHM$ with an amplitude $A$. What is the displacement of the particle when its potential energy is half of its total energy?

In simple harmonic motion,the total mechanical energy of a given system is $E$. If the mass of the oscillating particle $P$ is doubled,then the new energy of the system for the same amplitude is

Show that for a particle in linear $SHM$,the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

$A$ particle is executing simple harmonic motion with a time period $T$. At time $t = 0$,it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like:

Vedclass Test Series

Exam Paper Generator

Online Exam Module