An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at by an external force of . Determine the position of the mass at any time . Use as the acceleration due to gravity. Pay close attention to the units.

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

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