A 0.560-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.560-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.4 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 9.40 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 9.40 cm from the equilibrium position.
cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 5.40 cm.
s

Explanation / Answer

a) Lets Solve for maximum speed:

v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.114)^2 - (0)^2}(8/0.560)]
v = 0.43 m/sec

b) Lets Solve for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.560)(0.114)
a = - 1.628 m/sec^2

c) v = ?, when X = 0.094 m
v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.114)^2 - (0.094^2)}8/0.560]
v = 0.243 m/sec

d)a = ?, when X = 0.094 m
a = - (k/m)X
a = - (8/0.560)(0.094)
a = - 1.34 m/sec^2

e) T = time interval for object to move from x = 0 to x = 5.40cm
a = -(k/m)X
a = -(8/0.56)(0.0540)
a = - 0.77m/sec^2

T = [-(4^2)X/a]
T = [-(4^2)0.054/-0.77
T = 3.32sec

Time interval required for the object to move from x = 0 to x = 5.40 cm. is T = 3.32sec

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