Answer to Question #126300 in Differential Equations for jse

Question

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

Answer 1

Answer

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

Workings

Conditions given:

Mass (m):

m = 250g

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

= 0.25kg

Spring Extension (L):

L = 1.568 cm

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

= 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

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

= 0.20ms^-1

Finding the spring constant (k)

$mg - kL = 0$ , 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.

=> $mg = kL$

k = $\frac {mg} {L}$

= $\frac {0.025 × 9.8} {0.01568}$

= $\frac {2.45} {0.01568}$

= 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 = $\frac {k} {m}$

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

Since, w₀² = $\frac {k} {m}$

=> w_{0 =} $\sqrt {\frac {156.25} {0.25} }$

= $\sqrt {626}$

= 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 = $\frac {0.20} {25}$

= 0.008

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