Question

The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic and potential energy are equal?

Answer 1

Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
$v_{max} = \omega A$
Therefore
$v(t) = -V_{max} sin(\omega t)$

The KE (kinetic energy) is given by
$KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)$

The PE (potential energy) is given by
$PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)$

When the KE and PE are equal, then
$v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)$

For the oscillating spring,
$\omega ^{2} = \frac{k}{m} \\ V_{max} = \omega A = \sqrt{ \frac{k}{m} } A$
Therefore
$v^{2} = \frac{k}{m} \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max} \,cos( \sqrt{ \frac{k}{m} t} )$

Answer: $v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )$