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. (Assume the position of the object is at the origin at

t = 0.)

(a) Calculate the maximum value of its speed.

(b) Calculate the maximum value of its acceleration.

cm/s2

(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/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 2.60 cm.

p.s can you please answer this question in detail showing every step/formula/diagram

Explanation / Answer

a) maximum speed:

v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.106)^2 - (0)^2}(8/0.460)]
v = 0.651 m/sec

b) maximum acceleration:
a = - (k/m)Xo
a = - (8/0.460)(0.106)
a = - 1.843 m/sec^2

c) v = ?, when X = 0.086 m
v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.106)^2 - (0.086^2)}8/0.460]
v = 0.576 m/sec

d) a = ?, when X = 0.086 m
a = - (k/m)X
a = - (8/0.460)(0.086)
a = - 1.49 m/sec^2

e) T = time interval for object to move from x = 0 to x = 2.60cm
a = -(k/m)X
a = -(8/0.460)(0.0260)
a = - 0.452m/sec^2

T = ?[-(4?^2)X/a]
T = ?[-(4?^2)0.0260/-0.452
T = 2.24 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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