A mass–spring–damper is described by y″ + 2.4y′ + 4y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
Differential equations worksheet with answers
1.Easy
y″ + 2.4y′ + 4y = 0
ζ =
2.Easy
A mass–spring–damper is described by y″ + 2.4y′ + 16y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
y″ + 2.4y′ + 16y = 0
ζ =
3.Medium
A can of soda at 6 °C is taken out of the fridge into a room at 28 °C. After 23 min its temperature is 25 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). Find the constant k in min⁻¹.
T0 = 6 °CTa = 28 °CT(23) = 25 °C
k = min⁻¹
4.Medium
Solve y″ + 2y′ + y = 0 with y(0) = 2 and y′(0) = 1. Find y(2).
y″ + 2y′ + y = 0y(0) = 2y′(0) = 1
y(2) =
5.Hard
Solve y′ + 3y = 10e−2t with y(0) = 3. Find y(1.5).
y′ + 3y = 10e−2ty(0) = 3
y(1.5) =
6.Hard
A system obeys y″ + 3y′ + 27.25y = 272.5 for t ≥ 0, starting from rest: y(0) = 0, y′(0) = 0. Find the peak value ymax of its step response.
y″ + 3y′ + 27.25y = 272.5y(0) = 0y′(0) = 0
ymax =
Answer key
- 1. ζ = 0.6
- 2. ζ = 0.3
- 3. k = 0.08663 min⁻¹
- 4. y(2) = 1.083
- 5. y(1.5) = 0.4201
- 6. ymax = 13.9
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