A mass–spring–damper is described by y″ + 6.4y′ + 64y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
Differential equations worksheet with answers
1.Easy
y″ + 6.4y′ + 64y = 0
ζ =
2.Easy
A mass–spring–damper is described by y″ + 4.8y′ + 36y = 0. Compare it with y″ + 2ζωny′ + ωn²y = 0 and find the damping ratio ζ.
y″ + 4.8y′ + 36y = 0
ζ =
3.Medium
Solve y″ + 8y′ + 16y = 0 with y(0) = 5 and y′(0) = 2. Find y(0.5).
y″ + 8y′ + 16y = 0y(0) = 5y′(0) = 2
y(0.5) =
4.Medium
A forged steel part at 320 °C is left to cool in a workshop at 27 °C. It follows Newton's law of cooling, dT/dt = −k·(T − Tₐ). With k = 0.02 min⁻¹, find its temperature T after 19 min.
T0 = 320 °CTa = 27 °Ck = 0.02 min⁻¹t = 19 min
T = °C
5.Hard
Solve y″ + 3y′ + 27.25y = 0 with y(0) = 1 and y′(0) = 0 (released from rest). Find y(0.5); angles in radians.
y″ + 3y′ + 27.25y = 0y(0) = 1y′(0) = 0
y(0.5) =
6.Hard
Solve y′ + 0.5y = 8e−3t with y(0) = 5. Find y(1.5).
y′ + 0.5y = 8e−3ty(0) = 5
y(1.5) =
Answer key
- 1. ζ = 0.4
- 2. ζ = 0.4
- 3. y(0.5) = 2.165
- 4. T = 227.4 °C
- 5. y(0.5) = −0.2936
- 6. y(1.5) = 3.838
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