A mass–spring–damper is described by y″ + 2.5y′ + 25y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
Differential equations worksheet with answers
1.Easy
y″ + 2.5y′ + 25y = 0
ζ =
2.Easy
A mass–spring–damper is described by y″ + 8y′ + 25y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
y″ + 8y′ + 25y = 0
ζ =
3.Medium
Solve y″ + 6y′ + 9y = 0 with y(0) = 1 and y′(0) = -2. Find y(1).
y″ + 6y′ + 9y = 0y(0) = 1y′(0) = -2
y(1) =
4.Medium
Solve y″ + 5y′ + 6y = 0 with y(0) = 1 and y′(0) = 6. Find y(2).
y″ + 5y′ + 6y = 0y(0) = 1y′(0) = 6
y(2) =
5.Hard
A system obeys y″ + 2y′ + 5y = 25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 2y′ + 5y = 25y(0) = 0y′(0) = 0
ymax =
6.Hard
A system obeys y″ + 3y′ + 11.25y = 11.25 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find y(0.5); angles in radians.
y″ + 3y′ + 11.25y = 11.25y(0) = 0y′(0) = 0
y(0.5) =
Answer key
- 1. ζ = 0.25
- 2. ζ = 0.8
- 3. y(1) = 0.09957
- 4. y(2) = 0.145
- 5. ymax = 6.039
- 6. y(0.5) = 0.731
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