Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How to solve this problem- 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:

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:

 

  • Option 1)

    \frac{A}{2}

  • Option 2)

    A

  • Option 3)

    \frac{A}{2\sqrt{2}}

  • Option 4)

    \frac{A}{\sqrt{2}}

Answers (1)

best_answer

 

Kinetic energy in S.H.M. -

K.E.= \frac{1}{2}mu^{2}\\ \: \: \: = \frac{1}{2}m\left ( A^{2} -x^{2}\right )\omega ^{2}

 

- wherein

K.E.= \frac{1}{2}K\left ( A^{2}-x^{2} \right ) \\K= m\omega ^{2}

U=\frac{1}{2}k x^{2}

K.E.=\frac{1}{2}k (A^{2}-x^{2})

U=k

x=\pm \frac{A}{\sqrt2}

 

 

Potential energy in S.H.M. -

P.E.= \frac{1}{2}Kx^{2}

 

- wherein

Where K= m\omega ^{2}

 

 


Option 1)

\frac{A}{2}

Option 2)

A

Option 3)

\frac{A}{2\sqrt{2}}

Option 4)

\frac{A}{\sqrt{2}}
Posted by

admin

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question