Question #126300
A mass of 250 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 20 cm/s, and if there is no damping, determine the position u of the mass at any time t.


Enclose arguments of functions in parentheses. For example, sin(2x).
Assume g = 9.8 m/s^2. Enter an exact answer.

u(t) = _____ m
1
Expert's answer
2020-07-20T18:28:21-0400

Answer

u(t) = 0.008Sin(25t) m


Workings


Conditions given:

Mass (m):

m = 250g

= 250g1000g\frac {250g} {1000g}

= 0.25kg


Spring Extension (L):

L = 1.568 cm

= 1.568cm100cm\frac {1.568cm} {100cm}

= 0.01568 m


Let u(t) be the position of the mass at any time, t seconds.

Then,

u(0) = 0 m, the mass is initially in equilibrium.


Velocity = u`(t)

u`(0) = 20cms-1

= 20cms1100cm\frac {20cms^{-1}} {100cm}

= 0.20ms-1


Finding the spring constant (k)

mgkL=0mg - kL = 0 , where m is the mass in kg, g is the acceleration due to gravity in ms-2 , k is the spring constant in Nm-1 and L is the spring extension in m.

=> mg=kLmg = kL


k = mgL\frac {mg} {L}


= 0.025×9.80.01568\frac {0.025 × 9.8} {0.01568}


= 2.450.01568\frac {2.45} {0.01568}


= 156.25 Nm-1


Differential Equation and solution

When there is no damping and forcing, the equation is given by:

mu`` + ku = 0


The solution to the equation is given by:


u(t) = ASin(w0t) + BCos(w0t), where w0 is the natural frequency of the system and

w02 = km\frac {k} {m}

A and B are determined by initial conditions and u(0) = B, u`(0) = Aw0


Since, w02 = km\frac {k} {m}


=> w0 = 156.250.25\sqrt {\frac {156.25} {0.25} }


= 626\sqrt {626}

= 25

Therefore,

u(t) = ASin(25t) + BCos(25t)


From initial conditions:

u(0) = ASin(25 × 0) + BCos(25 × 0)

0 = ASin(0) + BCos(0)

B = 0


Thus,

u(t) = ASin(25t) + [0 × Cos(25t)]

u(t) = ASin(25t)


=> u`(t) = 25ACos(25t)

u`(0) = 25ACos(25 ×0)

0.20 = 25ACos(0)

A = 0.2025\frac {0.20} {25}

= 0.008


So, u(t) = 0.008Sin(25t) m


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