A particle is executing SIIM with an amplitude of 0.2 m. At what distance from the mean position the potential energy of the particle will be equal to its kinetic energy ?

A particle is executing SIIM with an amplitude of 0.2 m. At what dista

1.	A particle is executing SIIM with an amplitude of 0.2 m. At what distance from the mean position the potential energy of the particle will be equal to its kinetic energy ?
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :P.E. at the displacement x <br/> `P.E.= (1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)m^(2)omega^(2)x^(2)`…….(1) <br/> K.E. at the displacement x is given by <br/> `K.E.= (1)/(2)m^(2)omega^(2)(A^(2)-x^(2))` …………(2) <br/> But PE=KE i.e., `(1)/(2)m^(2)omega^(2)x^(2)= (1)/(2)momega^(2)(A^(2)-x^(2))` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> x^(2)= A^(2)-x^(2)` or `implies <a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a>^(2)=A^(2) x^(2)= (A^(2))/(2)` <br/> `:. x= (A)/(sqrt(2)).` But A=0.2m <br/> `:. x= (0.2)/(sqrt(2))= 0.1 xx sqrt(2)= 0.1414m`</body></html>