A particle executes simple pendulum harmonic motion of amplitude A. at what distance from the mean position is its kinetic energy to its potential energy?

  • Option 1)

    0.51 A

  • Option 2)

    0.61 A

  • Option 3)

    0.71 A

  • Option 4)

    0.81 A

 

Answers (1)

As we discussed in concept

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}

 

 K=U

=>\frac{1}{2}mw^{2}(A^{2}-x^{2})=\frac{1}{2}mw^{2}x^{2}

\therefore A^{2}-x^{2}=x^{2}

=> A^{2}=2x^{2}

x=\frac{A}{\sqrt{2}}

=> x=0.71A

 


Option 1)

0.51 A

Option is incorrect

Option 2)

0.61 A

Option is incorrect

Option 3)

0.71 A

Option is correct

Option 4)

0.81 A

Option is incorrect

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions