Question

A 0.560-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.4 cm. (Assume the position of the object is at the origin at

* t* = 0.)

(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 9.40 cm
from the equilibrium position.

cm/s

(d) Calculate the value of its acceleration when the object is 9.40
cm from the equilibrium position.

cm/s^{2}

(e) Calculate the time interval required for the object to move
from *x* = 0 to *x* = 5.40 cm.

s

Answer #1

a) Lets Solve for maximum speed:

v = √[{(Xo)^2 - (X^2)}k/m]

v = √[{(0.114)^2 - (0)^2}(8/0.560)]

**v = 0.43 m/sec**

b) Lets Solve for maximum acceleration:

a = - (k/m)Xo

a = - (8/0.560)(0.114)

**a = - 1.628 m/sec^2**

c) v = ?, when X = 0.094 m

v = √[{(Xo)^2 - (X^2)}k/m]

v = √[{(0.114)^2 - (0.094^2)}8/0.560]

**v = 0.243 m/sec**

d)a = ?, when X = 0.094 m

a = - (k/m)X

a = - (8/0.560)(0.094)

**a = - 1.34 m/sec^2**

e) T = time interval for object to move from x = 0 to x =
5.40cm

a = -(k/m)X

a = -(8/0.56)(0.0540)

**a = - 0.77m/sec^2**

T = √[-(4π^2)X/a]

T = √[-(4π^2)0.054/-0.77

**T = 3.32sec**

Time interval required for the object to move from *x* =
0 to *x* = 5.40 cm. is **T = 3.32sec**

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