A body of mass 200g 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.2:3
- B.1:2
- C.1:1
- D.2:5
Show answer and solution
Correct option: A
Solution
The spring force provides the centripetal force. If natural length is L₀ and extension is x, then the radius of circular motion is r = L₀ + x.
Setting spring force = centripetal force:
kx = mω²(L₀ + x)
12.5x = 0.2 × 25 × (L₀ + x)
12.5x = 5L₀ + 5x
7.5x = 5L₀
x/L₀ = 5/7.5 = 2/3
So the ratio of extension to natural length is 2:3.
Setting spring force = centripetal force:
kx = mω²(L₀ + x)
12.5x = 0.2 × 25 × (L₀ + x)
12.5x = 5L₀ + 5x
7.5x = 5L₀
x/L₀ = 5/7.5 = 2/3
So the ratio of extension to natural length is 2:3.
[JEE Main 2023 · 24 Jan · Shift 2]