Questions on modelling with differential equations

Question 1

In the oscillator model y'' + by' + cy = 0, state what the y' term and the y term each represent physically.

Accepted Answer

The y term is the restoring force pulling back towards equilibrium, its coefficient the stiffness; the y' term is resistance, draining energy in proportion to speed. B1 B1 for the two terms.

Question 2

Classify the damping in y'' + 8y' + 25y = 0, and state the frequency and decay rate of the motion.

Accepted Answer

Discriminant 64 − 100 = −36 < 0: light damping. The roots are −4 ± 3i, so the motion oscillates at angular frequency 3 inside a decaying envelope e^−4t: frequency from the imaginary part, decay from the real part. M1 for the discriminant, A1 for light damping, B1 for the frequency and decay rate.

Question 3

For y'' + by' + cy = 0 with b ≥ 0 and c > 0, state the condition on the discriminant b² − 4c for each of light, critical and heavy damping.

Accepted Answer

Light damping: b² − 4c < 0. B1. Critical damping: b² − 4c = 0. B1. Heavy damping: b² − 4c > 0. B1.

Question 4

Write down the general solution of x'' + ω²x = 0, and explain why the period of every solution is 2π/ω whatever the initial displacement and velocity are.

Accepted Answer

x = A cos ωt + B sin ωt. B1. The initial conditions fix A and B, which set the amplitude and the phase, but every term still oscillates at the same angular frequency ω: changing A or B rescales or shifts the wave without changing how often it repeats. B1. So the period 2π/ω depends only on the equation, not on how the motion starts. B1.

Question 5

A shock absorber obeys y'' + 10y' + 25y = 0. Classify the damping, give the general solution, and explain why this case is the design target for suspension.

Accepted Answer

Discriminant 100 − 100 = 0: critical damping, repeated root m = −5, solution y = (A + Bt)e^−5t. Critical damping returns the system to rest as fast as possible without oscillating (at most one crossing of the equilibrium); lighter damping bounces, heavier creeps. M1 for the discriminant, A1 for critical damping, A1 for the general solution, B1 for the design reason.

Question 6

Classify the damping in y'' + 5y' + 4y = 0 and describe the long-term motion.

Accepted Answer

Discriminant 25 − 16 = 9 > 0: heavy damping, real roots −1 and −4. The solution Ae^−t + Be^−4t creeps back to equilibrium without ever oscillating, the slower exponential setting the pace at late times. M1 for the discriminant, A1 for heavy damping with the roots, B1 for the long-term description.

Question 7

Solve the coupled system dx/dt = 2y, dy/dt = −2x with x(0) = 3 and y(0) = 0.

Accepted Answer

Differentiate the first equation: x'' = 2 dy/dt = −4x, so x'' + 4x = 0 and x = A cos 2t + B sin 2t. x(0) = 3 gives A = 3; y = x'/2 = −A sin 2t + B cos 2t at t = 0 gives B = 0. So x = 3 cos 2t and y = −3 sin 2t: the pair orbit a circle of radius 3 at angular speed 2. M1 for differentiating to eliminate y, A1 for x'' + 4x = 0, A1 for the general solution, M1 for applying both conditions, A1 for x and y.

Question 8

An oscillator obeys y'' + ky' + 25y = 0, where the resistance k is a positive constant that can be adjusted. Find the value of k for which the damping is critical, and classify the damping when k = 6 and when k = 26, describing the motion in each case.

Accepted Answer

Critical damping needs a zero discriminant: k² − 100 = 0, and since k is positive, k = 10. At k = 6 the discriminant is 36 − 100 = −64 < 0: light damping, roots −3 ± 4i, so the motion oscillates at angular frequency 4 inside the shrinking envelope e^−3t. At k = 26 it is 676 − 100 = 576 > 0: heavy damping, real roots −1 and −25, and the system creeps back to equilibrium without oscillating, the e^−t term setting the pace at late times. Increasing the resistance past the critical value makes the return slower, not faster. M1 for setting the discriminant to zero, A1 for k = 10, B1 for light damping at k = 6 with its description, B1 for heavy damping at k = 26 with its description.

Question 9

A particle moves along a straight line so that its displacement x metres from a point O at time t seconds satisfies x'' + 16x = 0, with x = 1.5 and x' = 8 when t = 0. Find x in terms of t, write it in the form R cos(4t − α), and state the amplitude and period of the motion and the particle's greatest speed.

Accepted Answer

The auxiliary equation m² + 16 = 0 has roots ±4i, so this is simple harmonic motion with x = A cos 4t + B sin 4t. x(0) = 1.5 gives A = 1.5. Differentiating, x' = −4A sin 4t + 4B cos 4t, and x'(0) = 8 gives 4B = 8, so B = 2. Hence x = 1.5 cos 4t + 2 sin 4t.
In harmonic form R = √(1.5² + 2²) = 2.5 and tan α = 2/1.5, so α = 0.927 and x = 2.5 cos(4t − 0.927).
The amplitude is 2.5 m and the period is 2π/4 = π/2 ≈ 1.57 s. Neither depends on solving for the phase: the amplitude needs both starting values, because the particle is released moving, not at an endpoint.
The greatest speed is Rω = 2.5 × 4 = 10 m/s, taken as the particle passes through O. M1 for the general solution from the auxiliary equation, A1 for A = 1.5 and B = 2, M1 for the harmonic form with R = √(A² + B²), A1 for the amplitude and period, B1 for the greatest speed.

Question 10

For y'' + 2y' + 10y = 3sin t, state whether the transient part of the general solution tends to zero as t → ∞, justifying your answer from the roots of the auxiliary equation, without finding the full solution.

Accepted Answer

The auxiliary equation m² + 2m + 10 = 0 has discriminant 4 − 40 = −36, so m = −1 ± 3i. M1 for the discriminant, A1 for the roots. Both roots have real part −1, which is negative, so the complementary function e^−t(A cos 3t + B sin 3t) does tend to zero as t → ∞. B1. Only then does the particular integral describe the long-term response; a root with real part zero or positive would leave the complementary function present forever, and the particular integral alone would not describe the motion correctly for large t. B1.

Question 11

Two quantities x and y satisfy dx/dt = −3x + 2y and dy/dt = 2x − 3y. Show that eliminating y gives x'' + 6x' + 5x = 0, and describe the long-term behaviour of x.

Accepted Answer

From the first equation, y = (x' + 3x)/2. M1 for rearranging. Differentiating the first equation, x'' = −3x' + 2y' = −3x' + 2(2x − 3y) = −3x' + 4x − 6y. Substituting 6y = 3(x' + 3x) = 3x' + 9x gives x'' = −3x' + 4x − 3x' − 9x, that is x'' + 6x' + 5x = 0. M1 for substituting, A1 for the equation. The auxiliary equation m² + 6m + 5 = (m + 1)(m + 5) = 0 has roots −1 and −5, both real and negative, so x = Ae^−t + Be^−5t decays to zero without oscillating. M1 for the auxiliary equation, A1 for the long-term description.

Question 12

A mass on a spring is driven so that its displacement y at time t satisfies y'' + 2y' + 5y = 8cos t, with y = 0 and y' = 0 when t = 0. Find y in terms of t, and describe the motion after a long time.

Accepted Answer

Complementary function. The auxiliary equation m² + 2m + 5 = 0 has discriminant 4 − 20 = −16, so m = −1 ± 2i and the complementary function is e^−t(A cos 2t + B sin 2t). Particular integral. Try y = p cos t + q sin t. Then y′ = −p sin t + q cos t and y″ = −p cos t − q sin t. Substituting and collecting: (−p + 2q + 5p)cos t + (−q − 2p + 5q)sin t = 8cos t, so 4p + 2q = 8 and 4q − 2p = 0. From the second, p = 2q; then 8q + 2q = 8 gives q = 0.8 and p = 1.6. General solution. y = e^−t(A cos 2t + B sin 2t) + 1.6cos t + 0.8sin t. Conditions. Apply them to the general solution, never to the complementary function alone. y(0) = 0 gives A + 1.6 = 0, so A = −1.6. Differentiating, y′(0) = −A + 2B + 0.8, so 1.6 + 2B + 0.8 = 0 and B = −1.2. Hence y = −e^−t(1.6cos 2t + 1.2sin 2t) + 1.6cos t + 0.8sin t. M1 for the auxiliary equation, A1 for the complementary function, M1 for a suitable particular integral, A1 for p and q, M1 for applying the conditions to the general solution, A1 for the value of A, A1 for the value of B, B1 for the steady state. Long term. The e^−t factor kills the first part, which is the transient. What remains is the steady state 1.6cos t + 0.8sin t: an oscillation at the driving frequency, of amplitude √(1.6² + 0.8²) = √3.2 ≈ 1.79.

Question 13

Two quantities x and y at time t satisfy the coupled equations dx/dt = −x + 2y and dy/dt = −2x − y, with x = 4 and y = 2 when t = 0. By reducing the pair to a single second order equation in x, find x and y in terms of t, and describe the long-term behaviour of the system.

Accepted Answer

Differentiate the first equation: x'' = −x' + 2 dy/dt = −x' + 2(−2x − y) = −x' − 4x − 2y. The first equation gives 2y = x' + x, so x'' = −x' − 4x − (x' + x), that is x'' + 2x' + 5x = 0.
The discriminant is 4 − 20 = −16, so the roots are −1 ± 2i and x = e^−t(A cos 2t + B sin 2t).
x(0) = 4 gives A = 4. Recover y from the first equation rather than integrating afresh: y = (x' + x)/2, so y(0) = 2 forces x'(0) = 0. Differentiating, x'(0) = −A + 2B = 0, so B = 2.
Hence x = e^−t(4 cos 2t + 2 sin 2t). Then x' = −10e^−t sin 2t, so y = (x' + x)/2 = e^−t(2 cos 2t − 4 sin 2t).
Long term, the factor e^−t sends both quantities to zero: the pair spiral in towards the origin, oscillating at angular frequency 2 inside a decaying envelope, the coupled analogue of light damping. M1 for differentiating the first equation, M1 for substituting 2y = x' + x, A1 for x'' + 2x' + 5x = 0, A1 for the general solution with roots −1 ± 2i, M1 for applying both conditions through y = (x' + x)/2, A1 for x and y, B1 for the long-term description.

Question 14

A vibrating system obeys y'' + ky' + 26y = 0, where k ≥ 0 is the resistance coefficient. Find the range of values of k for which the damping is light, and describe what happens to the motion as k increases towards the top of this range.

Accepted Answer

Light damping needs a negative discriminant: k² − 4(26) < 0, that is k² < 104. M1 for the discriminant, A1 for the inequality. Since k ≥ 0, the range is 0 ≤ k < 2√26. A1 for the range. The roots are m = (−k ± √(k² − 104))/2, oscillating at angular frequency √(104 − k²)/2 inside the envelope e^−kt/2. B1 for the roots in this form. As k increases towards 2√26, k² − 104 approaches zero from below, so the frequency √(104 − k²)/2 falls towards zero. B1. The decay rate k/2 rises towards √26 at the same time, so the oscillations become slower and less prominent, and the motion approaches the critically damped case at k = 2√26 itself, where they stop entirely. B1.

Question 15

A damped oscillator's displacement follows y'' + 4y' + 29y = 0. Find the period of the oscillation, and the time taken for the amplitude of the oscillation to halve, giving each to 3 significant figures.

Accepted Answer

The auxiliary equation m² + 4m + 29 = 0 has discriminant 16 − 116 = −100, so m = −2 ± 5i. M1 for the discriminant, A1 for the roots. The general solution is y = e^−2t(A cos 5t + B sin 5t): oscillation at angular frequency 5 inside the envelope e^−2t. The period is 2π/5 = 1.26 s to 3 significant figures. M1 for the period formula, A1 for the value. The envelope halves when e^−2t = 1/2, that is 2t = ln 2, so t = (ln 2)/2 = 0.347 s to 3 significant figures. M1 for the halving equation, A1 for the time.