# Analysis of motion of block attached to spring

A block is attached to a spring of constant $$k$$ and natural length $$\ell_0$$. One end of spring is fixed and block is given a velocity $$v_0$$. System is is horizontal plane. The maximum extension in spring is given as $$\ell_0/4$$. the question is to find the velocity of block at maximum extension of spring.

My textbook has used conservation of angular momentum about the fixed point since torque of all forces is $$0$$.

I had tried using centripetal force $$= mv^2/r$$ at the maximum position since it can be assumed to move in a circle of radius $$r = \ell_0+\ell_0/4$$. Along with conservation of mechanical energy.

However this method does not give the correct answer.

Can anyone please explain why my method is wrong?

• It is not necessary for the block to move in a circle! That only occurs when the length of the string is fixed. So using the notion of centripetal force would in principle by incorrect! Commented Jun 24, 2019 at 4:05

The block does not move in a circle with radius $$r=\frac{5}{4}\ell_0$$. This is clearly seen as the block starts in a position with radius $$\ell_0$$ but ends up at a point of maximum extension where the radius is $$\frac{5}{4}\ell_0$$.
Assuming that the local behavior of the mass around the point of maximum extension is centripetal is not valid. We can see this as the magnitude of force changes as a consequence of Hooke's law and the fact that the length of the spring changes from its maximal length. That is, $$F(t_0\pm\textrm{d}t)\neq F(t_0)$$, where $$F(t_0)$$ is the force on the mass in the radial direction at the time $$t=t_0$$, when the spring is at maximum extension. In centripetal motion, we always have $$F(t\pm\textrm{d}t)=F(t)=\frac{mv^2}{r}$$.