Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, In a simple harmonic oscillator, at the mean position

In a simple harmonic oscillator, at the mean position

  • Option 1)

    kinetic energy is minimum, potential energy is maximum

  • Option 2)

    both kinetic and potential energies are maximum

  • Option 3)

    kinetic energy is maximum, potential energy is minimum

  • Option 4)

    both kinetic and potential energies are minimum.

 

Answers (1)

best_answer

As we learnt in

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}

 

 

Potential energy in S.H.M. -

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

 

- wherein

Where K= m\omega ^{2}

 

 

Hence, K.E. maximum and P.E. is minimum at mean position.

Correct option is 3.

 

 


Option 1)

kinetic energy is minimum, potential energy is maximum

This is an incorrect option.

Option 2)

both kinetic and potential energies are maximum

This is an incorrect option.

Option 3)

kinetic energy is maximum, potential energy is minimum

This is the correct option.

Option 4)

both kinetic and potential energies are minimum.

This is an incorrect option.

Posted by

divya.saini

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

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question