Question

A body of mass $m$ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is $1\, cm$. If the angular velocity is doubled, the elongation in the string is $5\, cm$. The original length of the spring is

• A. 16 cm
• B. 15 cm
• C. 14 cm
• D. 13 cm

Answer 1

Answer: (B)

Let the length of the spring is $l$.
When the system is whirled round in a horizontal circle the centripetal force is given by
$F = \frac{ mv^2 }{ r } = \frac{ m ( r \, \omega)^2 }{ r } = mr \omega^2 $
Then, $r = l +$ elongation
Given : elongation $= 1\, cm$ (in the first case)
For angular velocity $\omega$ the force required is
$F_1 = m ( l + 1) \omega^2 = kx = k \times 1 = k$
or $ k = m ( l + 1) \omega^2 $...(i)
For second case,$ \omega'= 2 \omega$ , elongation $= 5\, cm = x$ Radius, $r = l + s$
So, $ F_2 = m ( 1+ 5 ) ( 2 \omega)^2 = kx = k \times 5 = 5 k $
or $5k = 4 m ( l + 5) \omega^2 $ ..(ii)
Now, dividing Eq. (i) by Eq. (ii), we get
$ \frac{ k }{ 5 k } = \frac{ m ( l + 1) \omega^2 }{ 4 m ( l + 5) \omega^2 } $
$\Rightarrow 5 ( l + 1) = 4 ( l + 5) $
$\Rightarrow 5l + 5 = 4l + 20 $
$l = 20 - 5 = 15\, cm$