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 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/s^{2}

(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/s^{2}

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

s

Answer #1

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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