Question

A mass of 2 kg attached to a spring is vibratedhorizontally by displacing the mass 40 cm fromits equilibrium position and releasing it. Find themaximum velocity ofthe mass. The spring constant
is 24.5 N/m.​

Answer 1

Explanation:

Maximum velocity will be achieved when potential energy is zero or x=0 . In other words all potential energy will be converted to kinetic energy .

so

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2} \\ {v}^{2} = \frac{k {x}^{2} }{m} = \frac{24.5 \times {( \frac{40}{100}) }^{2} }{2} \\ v = 0.4\sqrt{12.25} = 0.4 \times 3.5 = 1.4 \ \frac{m}{ {s}^{2} }$

Answer 2

here is your answer in the attachment