Saved Bookmarks
| 1. |
A body describes SHM with an amplitude of 0.05 m and a period of 0.2 s. Find the acceleration and velocityof the body when the displacement is (a) 5cm (b) 3 cm (c ) zero cm. |
|
Answer» Solution :Given, `A=0.05m, T = 0.2s` We know that `v=omega sqrt(A^(2)-y^(2))` and `a=-omega^(2)y` Hence (i) for `y=0.05m, omega=(2PI)/(T)=(2xx3/142)/(0.2)="31.42 rads"^(-1)` and `omega^(2)=(31.42)^(2)=987.2` `v=31/42sqrt(5^(2)-5^(2))xx10^(-2)=-49.36ms^(-2)` (ii) for `y=0.03m, v=31.42xx10^(-2)sqrt(5^(2)-3^(2))=1.2568ms^(-1)` `a=-987.2xx0.03=-29.616ms^(-2)` (iii) `y=0, a=0` `v=31.42sqrt(5^(2))xx10^(-2)=1.571ms^(-1)`. |
|