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 :

Question

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 : (A) 1: 1 (B) 2: 3 (C) 2:5 (D) 1: 2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2b1ec8fb0f0f216e348e97ba8bf8b26a-.png" /> kx = m ( L 0+ x ) ω2 ⇒ 12.5 x =(1/5)( L 0+ x ) 25 ⇒ 1.5 x = L 0 ⇒ ( x / L 0)=(2/3)

$1 : 1$

$2 : 3$

$2 : 5$

$1 : 2$

$k x = m (L_{0} + x) ω^{2}$
$\Rightarrow 12.5 x = \frac{1}{5} (L_{0} + x) 25 \Rightarrow 1.5 x = L_{0}$
$\Rightarrow \frac{x}{L _{0}} = \frac{2}{3}$

1:1

2:3

2:5

1:2

kx=m(L0​+x)ω2 ⇒12.5x=51​(L0​+x)25⇒1.5x=L0​ ⇒L0​x​=32​

$1 : 1$

$2 : 3$

$2 : 5$

$1 : 2$

$k x = m (L_{0} + x) ω^{2}$
$\Rightarrow 12.5 x = \frac{1}{5} (L_{0} + x) 25 \Rightarrow 1.5 x = L_{0}$
$\Rightarrow \frac{x}{L _{0}} = \frac{2}{3}$