A 10kg crate sits on a distant planet. The planet has the same mass as Earth, but 80% of the radius of Earth (Earth radius = 637
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Answer:
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.
