Question

# A 0.560-kg object attached to a spring with a force constant of 8.00 N/m vibrates in...

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 12.6 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 10.60 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 10.60 cm from the equilibrium position.
cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 4.60 cm.
s

Consider x(t) = Asin(?t - ?), and we can take ? = 0 here.
Then v(t) = A?cos(?t)
and a(t) = -A?²sin(?t)
Also, ? = ?(k/m) = ?(8kg/s² / 0.56kg) = 3.78 rad/s

a) vmax = A? = 12.6cm * 3.78rad/s = 47.6 cm/s
b) amax = A?² = 12.6cm * (3.78rad/s)² = 180 cm/s² = 1.8 m/s²
c) When is the object 10.6 cm from the equilibrium position?
x(t) = 10.6 cm = 12.6cm * sin(3.78t)
3.78t = arcsin(10.6/12.6) = 0.9996 rads
t = 0.264 s
Then v(0.264) = 12.6cm * 3.78rad/s * cos(3.78* 0.264) = 25.8 cm/s

d) and a(0.264) = -12.6cm * (3.78rad/s)² * sin(3.78 * 0.264) = ±151.3 cm/s
depending on whether it is coming or going

e) By setting ? = 0, we've set x(0) = 0.
x(t) = 4.6 cm = 12.6cm * sin(3.78t)
3.78t = arcsin(4.6/12.6) = 0.374 rads
t = 0.09887 s = 98.87 ms

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