Saved Bookmarks
| 1. |
Determine the period of small longitudinal oscillations of a body with mass `m` in the system shown in figure. The stiffness values of the springs are `x_(1)` and `x_(2)`. The friction and the masses of the springs are negligible. |
|
Answer» The net unbalanced force on the block `m` when it is at a small horiozontal distance `x` from the equilibrium position becomes `(k_(1)+k_(2))x` From `F_(x)=mw_(x)` for the block `:` `-(k_(1)+k_(2))x=m ddot(x)` Thus` ddot (x)=-((k_(1)+k_(2))/(m))x` Hence the sought time period `T=2pi sqrt((m)/(k_(1)+k_(2)))` Alternate `:` Let us set the block `m` in motion to perform small oscillation . Let us locate the block when it is at a distance `x` from its equilibrium position. As the spring force is restoring conservative force and deformation of both the springs are same, so from the conservation of mechanical energy of the spring `-` block system `:` `(1)/(2)m ((dx)/(dt))^(2)+(1)/(2) k_(1)x^(2)+(1)/(2)k_(2)x^(2)=` Constant Differentiating with respect to time `(1)/(2) m 2 ddot(x)ddot(x)+(1)/(2)(k_(1)+k_(2))2xdot(x)=0` `ddot(x)=-((k_(1)+k_(2)))/(m)x` Hence the sought time period `T=2pisqrt((m)/(k_(1)+k_(2)))` |
|