Oscillations Question 2
Question 2 - 24 January - Shift 2
A body of mass $200 g$ is tied to a spring of spring constant $12.5 N / m$, while the other end of spring is fixed at point $O$. If the body moves about $O$ in a circular path on a smooth horizontal surface with constant angular speed $5 rad / s$, then the ratio of extension in the spring to its natural length will be :
(1) $1: 2$
(2) $1: 1$
(3) $2: 3$
(4) $2: 5$
Show Answer
Answer: (3)
Solution:
Formula: Force of a spring
Natural length $=L_0$
Extension $=\mathbf{x}$
$kx=m(L_0+x) \omega^{2}$
$\Rightarrow 12.5 x=\frac{1}{5}(L_0+x) 25 \Rightarrow 1.5 x=L_0$
$\Rightarrow \frac{x}{L_n}=\frac{2}{3}$
