A 0.460-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.460-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.0 cm.

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

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

(c) Calculate the value of its speed when the object is 10.00 cm from the equilibrium position.

Can you determine a time when the object is at this position? You can then use this time in your equation of motion. cm/s

(d) Calculate the value of its acceleration when the object is 10.00 cm from the equilibrium position.
cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 6.00 cm.
s

Explanation / Answer

w = sqrt(k/m) = 4.17 s^-1

a) Vmax = A*w = 12*4.17 = 50.04 cm /s

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b) a = w^2*A = 208.66 cm /s^2

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c) v = w*sqrt(A^2-x^2)

v = 4.17*sqrt((12*12)-(10*10)) = 27.66 cm /s

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d) a = w^2*x = 4.17*4.17*10 = 173.889 cm /s^2

e) x = A*sinwt

6 = 12*sin4.17t

t = 0.1256s

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