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 mass is made to rotate at an angular velocity of 2 rad/s. Find the elongation of the spring.

• A. 1 cm
• B. 4 cm
• C. 7 cm
• D. 10 cm

Answer 1

The correct option is A 1 cm

We know that,

$kl=m{v}^2/r=m{\omega }^{2}r=m{\omega }^2(l_0+l)$

Here $\omega$ is the angular velocity $l_0$ is the natural length $\left(0.5\right)m$ and $(l_0+l)$

Thus $(k-m{\omega }^2)l=m{\omega }^2{l}_{0}l=\frac{m{\omega }^2{l}_0}{k-m{\omega }^2}$

Putting value

$l=\frac{0.5\times 4\times 0.5}{100-0.5\times 4}m=\frac{1}{100m}=1cm$