(C) Given: Mass $m = 10 \ g$,Amplitude $a = 10 \ cm$,Time period $T = 2 \ s$,Displacement $y = 5 \ cm$.Angular frequency $\omega = \frac{2\pi}{T} = \f

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

(C) Given: Mass $m = 10 \ g$,Amplitude $a = 10 \ cm$,Time period $T = 2 \ s$,Displacement $y = 5 \ cm$.Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi \ rad/s$.The kinetic energy $K$ of a particle in simple harmonic motion is given by $K = \frac{1}{2} m \omega^2 (a^2 - y^2)$.Substituting the values: $K = \frac{1}{2} \times 10 \times (\pi)^2 \times (10^2 - 5^2)$.$K = 5 \times \pi^2 \times (100 - 25)$.$K = 5 \times \pi^2 \times 75$.$K = 375 \pi^2 \ erg$.

Class 11 Physics Oscillations Energy of Simple Harmonic Motion

Medium

$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

A
$37.5 \pi^2 \ erg$
B
$3.75 \pi^2 \ erg$
C
$375 \pi^2 \ erg$
D
$0.375 \pi^2 \ erg$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Oscillations

Subtopic

Energy of Simple Harmonic Motion

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

MediumJEE MAIN 2023

View Solution

The displacement between the position of maximum potential energy and the position of maximum kinetic energy for a particle executing $S.H.M.$ is

EasyAIPMT 2002

View Solution

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

Easy

View Solution

$A$ particle executing simple harmonic motion has a kinetic energy $K = K_0 \cos^2(\omega t)$. The maximum values of the potential energy and the total energy are respectively:

MediumAIPMT 2007

View Solution

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

Similar Questions

$A$ particle is executing simple harmonic motion $(SHM)$ of amplitude $A$ along the $x$-axis about $x = 0$. When its potential energy $(PE)$ equals kinetic energy $(KE)$,the position of the particle will be

The displacement between the position of maximum potential energy and the position of maximum kinetic energy for a particle executing $S.H.M.$ is

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

$A$ particle executing simple harmonic motion has a kinetic energy $K = K_0 \cos^2(\omega t)$. The maximum values of the potential energy and the total energy are respectively:

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module