Question

A 0.580-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.4cm.

(a) Calculate the maximum value of its speed.
_______________ cm/s

(b) Calculate the maximum value of its acceleration.
_______________ cm/s²

(c) Calculate the value of its speed when the object is 6.40 cm from the equilibrium position.
_______________ cm/s

(d) Calculate the value of its acceleration when the object is 6.40 cm from the equilibrium position.
________________ cm/s²

(e) Calculate the time interval required for the object to move from x = 0 to x = 2.40 cm.
_________________ s

Answer 1

Consider x(t) = Asin(?t - ?), and we can take ? = 0 here.
Then velocity,

dx(t)/dt = v(t) = A?cos(?t)
and acceleration,

d²x/dt² = a(t) = -A?

Answer 2

a) Solving for maximum speed:

v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.104)^2 - (0)^2}(8/0.58)]
v = 0.386 m/sec ANSWER

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.580)(0.104)
a = - 1.4344 m/sec^2

b) v = ?, when X = 0.064 m
v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.104)^2 - (0.064^2)}8/0.580]
v = 0.3044 m/sec ANSWER

a = ?, when X = 0.064 m
a = - (k/m)X
a = - (8/0.580)(0.064)
a = - 0.88275 m/sec^2 ANSWER

c) T = time interval for object to move from x = 0 to x = 2.4cm
a = -(k/m)X
a = -(8/0.58)(0.024)
a = - 0.331 m/sec^2

T = ?[-(4?^2)X/a]
T = ?[-(4?^2)0.024/-0.331
T = 1.69 sec ANSWER