InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
(a) A particle executes simple harmonnic motion with an amplitude of `10cm`. At what distance from the mean position are the kinetic and potential energies equal? (b) The maximum speed and acceleration of a particle executing simple harmonic motion are `10m//s` and `50cm//s`. Find the position(s) of the particle when the speed is `8cm//s`. |
|
Answer» (a) `K=U` `1/2momega^2(A^2-x^2)=1/2momega^2x^2` `A^2-x^2=x^2` `x^2=A^2/2` `x=+-A/sqrt2=+-(10)/(sqrt2)=+-5sqrt2cm` (b) `v_(max)=Aomega=10` (i) `a_(max)=Aomega^2=50` (ii) `(ii)//(i)impliesomega=5rad//s, A=2cm` `v=omegasqrt(A^2-x^2)` `8=5sqrt((2)^2-x^2)` `x^2=(2)^2-(1.6)^2=1.44` `x=+-1.2cm` |
|