Answer:
The speed of the mass halfway to the equilibrium position is 0.55 m/s.
Explanation:
if K is the kinetic energy and P is the potential energy then the total energy of the system is:
Etot = K + P
we given the amplitude, and the halfway distance is x = 1/2×A them we can get Pmax by:
Pmax = P + K
1/2×k×A^2 = 1/2×k×x^2 + 1/2×m×v^2
k×A^2 = k×x^2 + m×v^2
m×v^2 = k×(A^2 - x^2)
v^2 = k×(A^2 - x^2)/m
= (8.84)×((21.3×10^-2)^2 - (1/2(21.3×10^-2)^2))
= 0.30079647
v = 0.55 m/s
Therefore, the speed of the mass halfway to the equilibrium position is 0.55 m/s.