A 0.400 kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.400 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.0 cm.
(a) Calculate the maximum value (magnitude) of its speed and acceleration.
cm/s
cm/s2
(b) Calculate the speed and acceleration when the object is 6.00 cm from the equilibrium position.
cm/s
cm/s2
(c) Calculate the time interval required for the object to move from x = 0 to x = 6.00 cm.
s


Need to figure out part C. Part A is 53.7 cm/s and 240 cm/sec^2. Part B is 46.5 cm/sec and -120 cm/sec^2.

Thanks!

Explanation / Answer

a) Solving for maximum speed:

v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.12)^2 - (0)^2}(8/0.400)]
v = 0.54 m/sec ANSWER

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.400)(0.12)
a = - 2.4 m/sec^2

b) v = ?, when X = 0.06 m
v = [{(Xo)^2 - (X^2)}k/m]
v = [{(0.12)^2 - (0.06^2)}8/0.400]
v = 0.216 m/sec ANSWER

a = ?, when X = 0.06 m
a = - (k/m)X
a = - (8/0.400)(0.06)
a = - 1.2m/sec^2 ANSWER

c) T = time interval for object to move from x = 0 to x = 6.00cm
a = -(k/m)X
a = -(8/0.400)(0.0600)
a = - 1.2 m/sec^2

T = [-(4^2)X/a]
T = [-(4^2)0.0600/-1.2
T = 1.40 sec ANSWER

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.