Saved Bookmarks
| 1. |
A particle executes SHM with an amplitude 4 cm. Locate the position of point where is speed is half its maximum speed. At what displacement is potential energy equal to kinetic energy? |
|
Answer» Here, `a=4cm, V=omegasqrt(a^(2)-y^(2))` `(aomega)/(2)= omega sqrt(a^(2)-y^(2)) or (a^(2))/(4)=a^(2)-y^(2)` or `y^(2)=a^(2)-a^(2)//4=3a^(2)//4` or `y=sqrt(3)a//2 =sqrt(3)xx4//2=2sqrt(3)cm` As, `P.E.=K.E. ,` therefore, `(1)/(2)momega^(2)y^(2)=(1)/(2)momega^(2)(a^(2)-y^(2))` or `y^(2)=a^(2)-y^(2) or 2y^(2)=a^(2)` or `y=a//sqrt(2)=4//sqrt(2)=2sqrt(2)cm` |
|