A 0.600-kg object attached to a spring with a force constant of 8.00 N/m vibrate
ID: 1908518 • Letter: A
Question
A 0.600-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.2 cm. 1) Calculate the value of its speed when the object is 7.20 cm from the equilibrium position. ________________________cm/s 2) Calculate the value of its acceleration when the object is 7.20 cm from the equilibrium position. _________________________cm/s^2 3) Calculate the time interval required for the object to move from x = 0 to x = 5.20 cm. _________________________s.Explanation / Answer
A 0.450 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 13.0 cm.
(a) Calculate the maximum value (magnitude) of its speed and acceleration.
(b) Calculate the speed and acceleration when the object is 10.00 cm from the equilibrium position.
(c) Calculate the time interval required for the object to move from x = 0 to x = 7.00 cm.
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
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