Question

One end of a massless spring of spring constant 100 N/m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the table is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring(approximately).

• A. 1 cm
• B. 51 cm
• C. 50 cm
• D. None of these

Answer 1

The correct option is A

1 cm

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude $\frac{v^2}{r}$ . The horizontal force on the particle is due to the spring and equals kl, where l is the elongation and k is the spring constant. Thus, $kl=\frac{m{v}^2}{r}=m{\omega }^{2}r=m{\omega }^2(l_0+l).$
$kl=\frac{1}{2}\left(4\right)(l+l_0)$
$\Rightarrow 100l=2(l+\frac{1}{2})$
$\Rightarrow l=1cm$