Saved Bookmarks
| 1. |
A body of mass m falls from a height h on to the pan of a spring balance, figure, The masses of the pan and spring are negligible. The spring constant of the spring is k. The body gets attached to the pan and starts executing S.H.M. in the vertical direction. Find the amplitude and energy of oscillation. A. `(mg)/(k)sqrt(1+(2kh)/(mg))`B. `sqrt((k)/(mg))(1+(2kh)/(mg))`C. `sqrt((mg)/(k))(1+(2kh)/(mg))`D. `(k)/(mg)sqrt(1+(kh)/(mg))` |
|
Answer» Correct Answer - A When a body of mass `m` falls from height `h` onoto the pan of spring balance, let the spring of spring balance gets compressed by a length `x`. The loss of `P.E.` of the mass `=mg(h+x)`. The gain in elastic potential energy by the spring dur to compression `=(1)/(2)kx^(2)` According to law of conservation of energy, we have `mg(h+x)=(1)/(2)kx^(2)` or `(1)/(2)kx^(2)-mgx-mgh=0` or `x^(2)-(2mgx)/(k)-(2mgh)/(k)=0` Solving this quadric equation, we have `x=((2mg)/(k)+-sqrt(((2mg)/(k))^(2)+((8mgh)/(k))))/(2)` `=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` In equilibrium position, the spring will be compressed by distance `(mg)/(k)`. If `r` is the amplitude of oscillation then `(mkg)/(k) +r=(mg)/(k)+-(mg)/(k)sqrt(1+(2kh)/(mg))` `:.` amplitude, `r=(mg)/(k)sqrt(1+(2kh)/(mg))` |
|