Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Try this! A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is halfof the maximum speed, its displacement y is

A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is 

  • Option 1)

    \frac{A}{2}

  • Option 2)

    \frac{A}{\sqrt2}

  • Option 3)

    \frac{A\sqrt3}{2}

  • Option 4)

    \frac{2A}{\sqrt3}

 

Answers (2)

best_answer

\\*v_{max} = \omega A \Rightarrow v = \frac{\omega A}{2} = \omega\sqrt{A^{2}- y^{2}} \\*\Rightarrow A^{2} - y^{2} = \frac{A^{2}}{4} \Rightarrow y^{2} = \frac{3}{4}A^{2} \Rightarrow y = \frac{\sqrt{3}}{2}A

 

Mean Position -

A position  during oscillation where the particle is at equilibrium position, i.e. net force on particle at this position is zero.

- wherein

Force acting on particle always point towards mean position.

 

 

 


Option 1)

\frac{A}{2}

This is incorrect.

Option 2)

\frac{A}{\sqrt2}

This is incorrect.

Option 3)

\frac{A\sqrt3}{2}

This is correct.

Option 4)

\frac{2A}{\sqrt3}

This is incorrect.

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

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

A2

Posted by

Koli hitesh

View full answer

Similar Questions

Trending Articles/News

anam-khan

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

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question