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

The Correct Option is 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 $ $ \hspace20mm$ ..(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 $ $ \hspace20mm$ ..(ii)
Now, dividing E (i) by E (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