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 10.6 cm.

(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 8.60 cm from the equilibrium position. (----- cm/s ?)

(d) Calculate the value of its acceleration when the object is 8.60 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 ?)

Explanation / Answer

a) Solving for maximum speed:

v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.13)^2 - (0)^2}(8/0.450)]
v = 0.55 m/sec ANSWER

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.450)(0.13)
a = - 2.31 m/sec^2

b) v = ?, when X = 0.1000 m
v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.13)^2 - (0.1000^2)}8/0.450]
v = 0.350 m/sec ANSWER

a = ?, when X = 0.1000 m
a = - (k/m)X
a = - (8/0.450)(0.1000)
a = - 1.78 m/sec^2 ANSWER

c) T = time interval for object to move from x = 0 to x = 7.00cm
a = -(k/m)X
a = -(8/0.45)(0.0700)
a = - 1.24 m/sec^2

T = ?[-(4?^2)X/a]
T = ?[-(4?^2)0.0700/-1.24
T = 1.49 sec ANSWER

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