Question

A mass M is attached to a massless spring of spring constant K. The mass is moved through a distance x and released. The maximum velocity of the mass is:

• A. $\frac{2\pi M}{K}$
• B. $x\sqrt{\frac{k}{m}}$
• C. $\frac{\sqrt{K}}{M}$
• D. $\frac{\sqrt{K}}{Mx}$

Answer 1

The correct option is B $x\sqrt{\frac{k}{m}}$

Initial total energy of the system = KE +PE
initially KE is zero and PE is $(k\ast x^2)/2$
as we know, maximum velocity will be attained when the block is at the equilibrium position because at equilibrium net force on the block will be zero.
now, when the block passes through the equilibrium position
we have,
PE is zero as there is neither expansion nor contraction in the spring
$KE=(m\ast v^2)/2$
applying conservation of energy gives us
$(m\ast v^2)/2=(k\ast x^2)/2$
$v=\left(\sqrt{\left(k/m\right)}\right)\ast x$