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 A spring with a spring constant of 450 N/m is stretched 15 cm from its equilibrium position and released. a) If the mass attached to the spring is 2.5 kg, what is the frequency of the oscillation? b) What is the maximum kinetic energy of the mass? c) What is the maximum speed?

Respuesta :

Answer:

a)frequency of the oscillation = 1/(2*pi)*square root (k/x) =1/(2*pi)*square root (450/(15*10^-2))=8.72 cycle/second

b)spring potential energy = 0.5*k*(x)^2

=0.5*450*(15*10^-2)^2 =5.0625 joule

maximum kinetic energy =spring potential energy

c)

maximum kinetic energy=5.0625

kinetic energy=0.5*m*v^2

v=square toot ((5.0625/(0.5*2.5)) =2 m/s