Hard Further Maths Modelling with differential equations Questions

Challenging, exam-style Further Maths Modelling with differential equations questions with worked solutions. Stretch yourself on the hardest differential-equation-modelling, forced-oscillations, particular-integral, coupled-systems problems.

differential-equation-modellingforced-oscillationsparticular-integralcoupled-systemseliminationinitial-conditions
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A cam follower of mass 99 kg is held against its cam by a spring and a damper. Its displacement xx centimetres from the contact position at time tt seconds satisfies 9x¨+kx˙+4x=09\ddot{x}+k\dot{x}+4x=0, where k>0k>0 is a damping constant. Find the value of kk for which the system is critically damped.
Show worked solution

Worked solution

  1. Form the auxiliary equation

    9m2+km+4=09m^{2}+km+4=0

    Trying x=emtx=e^{mt} turns the model into a quadratic in mm whose coefficients involve kk.

  2. State the condition for critical damping

    b2−4ac=0b^{2}-4ac=0

    Critical damping is exactly the boundary case of a repeated root.

  3. Write down the discriminant

    k2−4×9×4=k2−144k^{2}-4\times9\times4=k^{2} - 144

    Read aa, bb and cc straight off the auxiliary equation.

  4. Set the discriminant to zero and solve

    k2=144 ⇒ k=12k^{2}=144\ \Rightarrow\ k=12

    The damping constant is positive, so the negative root is rejected.

  5. Substitute the critical value back into the auxiliary equation

    9m2+12m+4=09m^{2}+12m+4=0

    With this value of kk the quadratic should collapse to a perfect square.

  6. Factorise the auxiliary equation

    9(m+23)2=09\left(m+\frac{2}{3}\right)^{2}=0

    A perfect square is exactly what a repeated root means.

  7. Check the repeated root

    m=−23 (twice)m=- \frac{2}{3}\ \text{(twice)}

    Both roots coincide, which is the definition of critical damping.

  8. Write down the resulting motion

    x=(A+Bt)e−23tx=\left(A+Bt\right)e^{- \frac{2}{3}t}

    The extra factor of tt is what a repeated root contributes.

  9. Confirm the motion has no oscillation

    (A+Bt)e−23t=0 at most once\left(A+Bt\right)e^{- \frac{2}{3}t}=0\ \text{at most once}

    A linear factor has only one root, so the model crosses equilibrium at most once.

  10. Test a smaller damping constant

    k=6 ⇒ b2−4ac=−108<0k=6\ \Rightarrow\ b^{2}-4ac=-108<0

    Halving kk makes the discriminant negative, so the model would oscillate.

  11. Test a larger damping constant

    k=24 ⇒ b2−4ac=432>0k=24\ \Rightarrow\ b^{2}-4ac=432>0

    Doubling kk makes the discriminant positive, so the model would be heavily damped.

  12. Note what happens on either side of the critical value

    k<12 light,k>12 heavyk<12\ \text{light},\qquad k>12\ \text{heavy}

    Below the critical value the model oscillates; above it, it creeps back.

  13. Interpret the answer physically

    k=12 gives the fastest return to equilibriumk=12\ \text{gives the fastest return to equilibrium}

    Critical damping is what a designer wants when overshoot must be avoided.

  14. Recall the auxiliary equation

    ax¨+bx˙+cx=0 ⇒ am2+bm+c=0a\ddot{x}+b\dot{x}+cx=0\ \Rightarrow\ am^{2}+bm+c=0

    Trying x=emtx=e^{mt} turns the differential equation into a quadratic in mm.

  15. Recall how the discriminant classifies the damping

    b2−4ac>0 heavy,b2−4ac=0 critical,b2−4ac<0 lightb^{2}-4ac>0\ \text{heavy},\quad b^{2}-4ac=0\ \text{critical},\quad b^{2}-4ac<0\ \text{light}

    The nature of the roots decides whether the model oscillates on its way to rest.

  16. State the critical damping constant

    k=12k=12

    This is the unique positive value of kk that gives critical damping.

Answer
k=12k=12
Question 2
9 markschallenging
A precision balance of mass 55 kg is fitted with a weak damper. Its displacement xx millimetres from the balance position at time tt seconds satisfies 5x¨+2x˙+7x=05\ddot{x}+2\dot{x}+7x=0. Which of the following correctly describes the damping in this model?
Show worked solution

Worked solution

  1. Form the auxiliary equation

    5m2+2m+7=05m^{2}+2m+7=0

    The type of damping is decided entirely by the roots of this quadratic.

  2. Evaluate the discriminant

    b2−4ac=(2)2−4×5×7=−136b^{2}-4ac=\left(2\right)^{2}-4\times5\times7=-136

    The sign of the discriminant is the whole of the classification.

  3. Interpret the sign of the discriminant

    b2−4ac<0 ⇒ light dampingb^{2}-4ac<0\ \Rightarrow\ \text{light damping}

    Two distinct real roots give heavy damping, a repeated root gives critical damping and complex roots give light damping.

  4. Solve the auxiliary equation

    m=−(2)±−1362×5=−15+34i5, −15−34i5m=\frac{-\left(2\right)\pm\sqrt{-136}}{2\times5}=- \frac{1}{5} + \frac{\sqrt{34} i}{5},\ - \frac{1}{5} - \frac{\sqrt{34} i}{5}

    The roots confirm the classification made from the discriminant.

  5. Write down the complementary function

    x=e−15t(Acos⁡(345t)+Bsin⁡(345t))x=e^{- \frac{1}{5}t}\left(A\cos\left(\frac{\sqrt{34}}{5}t\right)+B\sin\left(\frac{\sqrt{34}}{5}t\right)\right)

    The form of the solution is the real content of the classification.

  6. Compare the damping coefficient with its critical value

    bcrit=2ac=25×7=235b_{\text{crit}}=2\sqrt{ac}=2\sqrt{5\times7}=2 \sqrt{35}

    This is the value of bb at which the discriminant would vanish.

  7. Form the damping ratio

    bbcrit=2235=3535\frac{b}{b_{\text{crit}}}=\frac{2}{2 \sqrt{35}}=\frac{\sqrt{35}}{35}

    A ratio below 11 is light damping, exactly 11 is critical and above 11 is heavy.

  8. Describe the resulting motion

    x crosses x=0 repeatedlyx\ \text{crosses}\ x=0\ \text{repeatedly}

    Complex roots give a genuine oscillation inside a decaying envelope.

  9. Confirm the motion decays

    Re⁡(m)<0 ⇒ x→0\operatorname{Re}\left(m\right)<0\ \Rightarrow\ x\rightarrow 0

    Every root has a negative real part, so the model does return to equilibrium.

  10. Reject the undamped option

    the coefficient of x˙ is 2≠0\text{the coefficient of}\ \dot{x}\ \text{is}\ 2\neq 0

    A non-zero x˙\dot{x} coefficient means there is genuine damping, so the roots cannot be purely imaginary.

  11. Reject the negative-damping option

    both roots have Re⁡(m)<0\text{both roots have}\ \operatorname{Re}\left(m\right)<0

    No root has a positive real part, so nothing in the solution can grow.

  12. Reject the two remaining classifications

    b2−4ac≠0 and b2−4ac<0b^{2}-4ac\neq 0\ \text{and}\ b^{2}-4ac<0

    The discriminant can only have one sign, so exactly one classification survives.

  13. Recall the auxiliary equation

    ax¨+bx˙+cx=0 ⇒ am2+bm+c=0a\ddot{x}+b\dot{x}+cx=0\ \Rightarrow\ am^{2}+bm+c=0

    Trying x=emtx=e^{mt} turns the differential equation into a quadratic in mm.

  14. Recall how the discriminant classifies the damping

    b2−4ac>0 heavy,b2−4ac=0 critical,b2−4ac<0 lightb^{2}-4ac>0\ \text{heavy},\quad b^{2}-4ac=0\ \text{critical},\quad b^{2}-4ac<0\ \text{light}

    The nature of the roots decides whether the model oscillates on its way to rest.

  15. Select the correct description

    b2−4ac=−136 ⇒ light dampingb^{2}-4ac=-136\ \Rightarrow\ \text{light damping}

    The discriminant settles the classification with no further work.

Answer
Light (under) damping: the auxiliary equation has complex roots, so the system oscillates with an exponentially decaying amplitude
Question 3
9 markschallenging
A lightly damped resonator is driven by an alternating field. Its displacement xx nanometres from the equilibrium position at time tt seconds satisfies x¨+2x˙+5x=10cos⁡(t)\ddot{x}+2\dot{x}+5x=10 \cos\left(t \right). Which of the following is the steady-state solution of this model?
Show worked solution

Worked solution

  1. Form the auxiliary equation

    m2+2m+5=0m^{2}+2m+5=0

    Substituting x=emtx=e^{mt} into the homogeneous equation turns it into a quadratic in mm.

  2. Evaluate the discriminant

    b2−4ac=−16b^{2}-4ac=-16

    Its sign is what decides the nature of the damping.

  3. Solve the auxiliary equation

    m=−1+2i, −1−2im=-1 + 2 i,\ -1 - 2 i

    The discriminant is negative, so there are a complex-conjugate pair of roots.

  4. Write down the complementary function

    x=e−t(Acos⁡(2t)+Bsin⁡(2t))x=e^{-t}\left(A\cos\left(2t\right)+B\sin\left(2t\right)\right)

    This is the standard form for a complex-conjugate pair of roots.

  5. Note that the complementary function is a transient

    e−t(Acos⁡(2t)+Bsin⁡(2t))→0 as t→∞e^{-t}\left(A\cos\left(2t\right)+B\sin\left(2t\right)\right)\rightarrow 0\ \text{as}\ t\rightarrow\infty

    The steady state is therefore the particular integral on its own.

  6. Substitute the trial function and compare coefficients

    λ=2,μ=1\lambda=2,\qquad\mu=1

    Matching the coefficients of cos⁡(t)\cos\left(t\right) and sin⁡(t)\sin\left(t\right) gives two equations.

  7. Differentiate the trial function twice

    x˙=−2sin⁡(t)+cos⁡(t),x¨=−sin⁡(t)−2cos⁡(t)\dot{x}=- 2 \sin\left(t \right) + \cos\left(t \right),\qquad\ddot{x}=- \sin\left(t \right) - 2 \cos\left(t \right)

    Both derivatives are needed before the trial function can be substituted.

  8. Substitute the fitted function into the model

    x¨+2x˙+5x=10cos⁡(t)\ddot{x}+2\dot{x}+5x=10 \cos\left(t \right)

    The residual vanishes identically, so this really is a solution of the forced model.

  9. Write the steady state in harmonic form

    R=λ2+μ2=5R=\sqrt{\lambda^{2}+\mu^{2}}=\sqrt{5}

    The amplitude of the long-term oscillation follows immediately from the two constants.

  10. Test a sign-flipped option by substitution

    sin⁡(t)−2cos⁡(t) ⇒ residual=8sin⁡(t)−16cos⁡(t)≠0\sin\left(t \right) - 2 \cos\left(t \right)\ \Rightarrow\ \text{residual}=8 \sin\left(t \right) - 16 \cos\left(t \right)\neq 0

    Flipping the sign of λ\lambda leaves a non-zero residual, so that option is wrong.

  11. Note that the transient is not part of the steady state

    e−t(Acos⁡(2t)+Bsin⁡(2t))→0e^{-t}\left(A\cos\left(2t\right)+B\sin\left(2t\right)\right)\rightarrow 0

    Any option containing an exponential factor could not be the steady state.

  12. Note the period of the steady state

    period=2π\text{period}=\frac{2\pi}{}

    In the long run the model is driven at the frequency of the forcing, not its own.

  13. Reject any option with λ\lambda and μ\mu interchanged

    λ≠μ\lambda\neq\mu

    The two constants play different roles, so swapping them changes the function.

  14. Recall the structure of the full solution

    x=complementary function+particular integralx=\text{complementary function}+\text{particular integral}

    The complementary function carries the arbitrary constants; the particular integral carries the forcing.

  15. Recall the trial particular integral for a sinusoidal forcing term

    x=λcos⁡(Ωt)+μsin⁡(Ωt)x=\lambda\cos\left(\Omega t\right)+\mu\sin\left(\Omega t\right)

    Both a cosine and a sine are needed because differentiating swaps them.

  16. Recall the trial particular integral for an exponential forcing term

    x=λeαtx=\lambda e^{\alpha t}

    An exponential reproduces itself under differentiation, so one constant suffices.

  17. Select the steady-state solution

    x=sin⁡(t)+2cos⁡(t)x=\sin\left(t \right) + 2 \cos\left(t \right)

    This is what the model settles down to once the transient has decayed.

Answer
sin⁡(t)+2cos⁡(t)\sin\left(t \right) + 2 \cos\left(t \right)
Question 4
9 markschallenging
An undamped instrument mount is driven at a frequency different from its own, starting from rest at the equilibrium position. Its displacement xx micrometres at time tt seconds satisfies x¨+9x=6cos⁡(2t)\ddot{x}+9x=6 \cos\left(2 t \right). When t=0t=0, x=0x=0 and x˙=0\dot{x}=0. Which of the following best describes the behaviour of the system as tt becomes large?
Show worked solution

Worked solution

  1. Form the auxiliary equation

    m2+9=0m^{2}+9=0

    Substituting x=emtx=e^{mt} into the homogeneous equation turns it into a quadratic in mm.

  2. Evaluate the discriminant

    b2−4ac=−36b^{2}-4ac=-36

    Its sign is what decides the nature of the damping.

  3. Solve the auxiliary equation

    m=−3i, 3im=- 3 i,\ 3 i

    The discriminant is negative, so there are a complex-conjugate pair of roots.

  4. Write down the complementary function

    x=Acos⁡(3t)+Bsin⁡(3t)x=A\cos\left(3t\right)+B\sin\left(3t\right)

    This is the standard form for a complex-conjugate pair of roots.

  5. Find the particular integral

    xPI=6cos⁡(2t)5x_{\text{PI}}=\frac{6 \cos\left(2 t \right)}{5}

    This is the part of the solution forced by the right-hand side.

  6. Form the full solution

    x=6cos⁡(2t)5−6cos⁡(3t)5x=\frac{6 \cos\left(2 t \right)}{5} - \frac{6 \cos\left(3 t \right)}{5}

    Applying the initial conditions to CF + PI gives the actual motion.

  7. Examine each part as t→∞t\rightarrow\infty

    no damping: nothing decays\text{no damping: nothing decays}

    With no damping term nothing in the complementary function decays.

  8. Note the sign of the real part of each root

    Re⁡(m)=0\operatorname{Re}\left(m\right)=0

    A negative real part makes that part of the solution decay.

  9. Compare the two frequencies present

    natural and forcing frequencies\text{natural and forcing frequencies}

    If they are equal and there is no damping the response grows without bound.

  10. Check the size of the solution at a large time

    x(200)≈0.5685x\left(200\right)\approx 0.5685

    Evaluating the solution far into the future confirms the classification.

  11. Test whether the solution settles to a single sinusoid

    x(t+T)≠x(t)x\left(t+T\right)\neq x\left(t\right)

    The response never repeats with the period of the forcing, so it does not settle.

  12. Compare the natural and the forcing frequencies

    ω0=3,Ω=2\omega_{0}=3,\qquad\Omega=2

    Equal frequencies with no damping is exactly the resonance condition.

  13. Note the role of the damping term

    coefficient of x˙=0\text{coefficient of}\ \dot{x}=0

    With no damping there is no transient to die away, so nothing settles.

  14. Recall the structure of the full solution

    x=complementary function+particular integralx=\text{complementary function}+\text{particular integral}

    The complementary function carries the arbitrary constants; the particular integral carries the forcing.

  15. Recall the trial particular integral for a sinusoidal forcing term

    x=λcos⁡(Ωt)+μsin⁡(Ωt)x=\lambda\cos\left(\Omega t\right)+\mu\sin\left(\Omega t\right)

    Both a cosine and a sine are needed because differentiating swaps them.

  16. Select the correct description

    t→∞:no damping: nothing decayst\rightarrow\infty:\quad \text{no damping: nothing decays}

    This is what the model predicts in the long run.

Answer
xx stays bounded but never settles down, because it is the sum of two oscillations with different periods
Question 5
9 markschallenging
A test rig of mass 33 kg is fitted with a damper and a spring so that the mass returns to rest in the shortest possible time. Its displacement xx metres from the equilibrium position at time tt seconds satisfies 3x¨+12x˙+12x=03\ddot{x}+12\dot{x}+12x=0. Which of the following correctly describes the damping in this model?
Show worked solution

Worked solution

  1. Form the auxiliary equation

    3m2+12m+12=03m^{2}+12m+12=0

    The type of damping is decided entirely by the roots of this quadratic.

  2. Evaluate the discriminant

    b2−4ac=(12)2−4×3×12=0b^{2}-4ac=\left(12\right)^{2}-4\times3\times12=0

    The sign of the discriminant is the whole of the classification.

  3. Interpret the sign of the discriminant

    b2−4ac=0 ⇒ critical dampingb^{2}-4ac=0\ \Rightarrow\ \text{critical damping}

    Two distinct real roots give heavy damping, a repeated root gives critical damping and complex roots give light damping.

  4. Solve the auxiliary equation

    m=−(12)±02×3=−2m=\frac{-\left(12\right)\pm\sqrt{0}}{2\times3}=-2

    The roots confirm the classification made from the discriminant.

  5. Write down the complementary function

    x=(A+Bt)e−2tx=\left(A+Bt\right)e^{-2t}

    The form of the solution is the real content of the classification.

  6. Compare the damping coefficient with its critical value

    bcrit=2ac=23×12=12b_{\text{crit}}=2\sqrt{ac}=2\sqrt{3\times12}=12

    This is the value of bb at which the discriminant would vanish.

  7. Form the damping ratio

    bbcrit=1212=1\frac{b}{b_{\text{crit}}}=\frac{12}{12}=1

    A ratio below 11 is light damping, exactly 11 is critical and above 11 is heavy.

  8. Describe the resulting motion

    x never crosses x=0 more than oncex\ \text{never crosses}\ x=0\ \text{more than once}

    Real roots give a sum of decaying exponentials, which changes sign at most once.

  9. Confirm the motion decays

    Re⁡(m)<0 ⇒ x→0\operatorname{Re}\left(m\right)<0\ \Rightarrow\ x\rightarrow 0

    Every root has a negative real part, so the model does return to equilibrium.

  10. Reject the undamped option

    the coefficient of x˙ is 12≠0\text{the coefficient of}\ \dot{x}\ \text{is}\ 12\neq 0

    A non-zero x˙\dot{x} coefficient means there is genuine damping, so the roots cannot be purely imaginary.

  11. Reject the negative-damping option

    both roots have Re⁡(m)<0\text{both roots have}\ \operatorname{Re}\left(m\right)<0

    No root has a positive real part, so nothing in the solution can grow.

  12. Reject the two remaining classifications

    b2−4ac=0 exactlyb^{2}-4ac=0\ \text{exactly}

    The discriminant can only have one sign, so exactly one classification survives.

  13. Recall the auxiliary equation

    ax¨+bx˙+cx=0 ⇒ am2+bm+c=0a\ddot{x}+b\dot{x}+cx=0\ \Rightarrow\ am^{2}+bm+c=0

    Trying x=emtx=e^{mt} turns the differential equation into a quadratic in mm.

  14. Recall how the discriminant classifies the damping

    b2−4ac>0 heavy,b2−4ac=0 critical,b2−4ac<0 lightb^{2}-4ac>0\ \text{heavy},\quad b^{2}-4ac=0\ \text{critical},\quad b^{2}-4ac<0\ \text{light}

    The nature of the roots decides whether the model oscillates on its way to rest.

  15. Select the correct description

    b2−4ac=0 ⇒ critical dampingb^{2}-4ac=0\ \Rightarrow\ \text{critical damping}

    The discriminant settles the classification with no further work.

Answer
Critical damping: the auxiliary equation has a repeated negative root, so the system returns to equilibrium as quickly as possible without oscillating

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