Question

2) A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.2 cm. (Assume the position of the object is at the origin at

t = 0.)

(a) Calculate the maximum value of its speed.
  cm/s

(b) Calculate the maximum value of its acceleration.
  cm/s²

(c) Calculate the value of its speed when the object is 8.20 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 8.20 cm from the equilibrium position.
Answer must be in cm/s²

Answer 1

angular velocity of system

w^2 =k/m = 8 / 0.52

w = 3.922 rad/s

a)

v = Aw = 3.922* 10.2 = 40 cm/s

b)

a = Aw^2 = 10.2* 3.922^2 = 156.897 m/s^2

c)

v^2 = w^2 ( A^2 - x^2)

v^2 = (8 / 0.52)( 10.2^2 - 8.2^2)

v = 23.794 cm/s

d)

A = kx/ m = 8* 8.2 / 0.52 = 126.15 cm/s^2

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