Further Maths Modelling with differential equations Practice Questions

Free Further Maths Modelling with differential equations practice questions with full step-by-step worked solutions. Covers differential-equation-modelling, simple-harmonic-motion, period, frequency. Practise exam-style problems and check your method.

differential-equation-modellingsimple-harmonic-motionperiodfrequencymaximum-speedamplitude
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A trolley on a smooth horizontal track is attached to a spring. Its displacement xx metres from the equilibrium position at time tt seconds satisfies x¨=16x\ddot{x}=-16x. Find the exact period of the oscillation.
Show worked solution

Worked solution

  1. Compare the model with the SHM equation

    x¨=ω2xwithω2=16\ddot{x}=-\omega^{2}x\quad\text{with}\quad\omega^{2}=16

    Reading off ω2\omega^{2} is the only information the period needs.

  2. Find the angular frequency

    ω=4\omega=4

    Take the positive square root; ω\omega is a rate, so it is positive.

  3. State the period

    T=π2T=\frac{\pi}{2}

    This is the exact time for one complete oscillation, in seconds.

Answer
T=π2T=\frac{\pi}{2}
Question 2
2 markseasy
A loudspeaker cone is driven by an alternating signal. Its displacement xx millimetres from rest at time tt seconds satisfies x¨+64x=5cos(ωt)\ddot{x}+64x=5 \cos\left(\omega t \right), where ω>0\omega>0 is the angular frequency of the forcing. Find the value of ω\omega at which resonance occurs.
Show worked solution

Worked solution

  1. Find the natural frequency of the undamped model

    1x¨+64x=0  x¨=641x1\ddot{x}+64x=0\ \Rightarrow\ \ddot{x}=-\frac{64}{1}x

    Setting the forcing term to zero leaves the free oscillation of the system.

  2. Read off the natural angular frequency

    ω02=641  ω0=8\omega_{0}^{2}=\frac{64}{1}\ \Rightarrow\ \omega_{0}=8

    This is the frequency at which the system oscillates when left alone.

  3. State the condition for resonance

    ω=ω0\omega=\omega_{0}

    There is no damping, so a forcing term at the natural frequency is already a solution of the homogeneous equation.

  4. State the resonant frequency

    ω=8\omega=8

    At this forcing frequency the amplitude of the response grows without bound.

Answer
ω=8\omega=8
Question 3
4 marksintermediate
A damped trolley on a shaking table is driven by the table. Its displacement xx centimetres from the equilibrium position at time tt seconds satisfies x¨+3x˙+2x=10sin(t)\ddot{x}+3\dot{x}+2x=10 \sin\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+3m+2=0m^{2}+3m+2=0

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

  2. Evaluate the discriminant

    b24ac=1b^{2}-4ac=1

    Its sign is what decides the nature of the damping.

  3. Solve the auxiliary equation

    m=1, 2m=-1,\ -2

    The discriminant is positive, so there are two distinct real roots.

  4. Write down the complementary function

    x=Aet+Be2tx=Ae^{-t}+Be^{-2t}

    This is the standard form for two distinct real roots.

  5. Note that the complementary function is a transient

    Aet+Be2t0 as tAe^{-t}+Be^{-2t}\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=3,μ=1\lambda=-3,\qquad\mu=1

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

  7. Select the steady-state solution

    x=sin(t)3cos(t)x=\sin\left(t \right) - 3 \cos\left(t \right)

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

Answer
sin(t)3cos(t)\sin\left(t \right) - 3 \cos\left(t \right)
Question 4
6 markshard
A hydraulic strut is modelled as a damped oscillator. The displacement hh centimetres of the piston from its rest position at time tt seconds satisfies 9h¨+6h˙+h=09\ddot{h}+6\dot{h}+h=0. Which of the following is the auxiliary equation for this model?
Show worked solution

Worked solution

  1. Try a solution of the form h=emth=e^{mt}

    h˙=memt,h¨=m2emt\dot{h}=me^{mt},\qquad\ddot{h}=m^{2}e^{mt}

    An exponential reproduces itself under differentiation, which is what makes this work.

  2. Substitute into the model

    (9m2+6m+1)emt=0\left(9m^{2}+6m+1\right)e^{mt}=0

    Every term carries the same factor emte^{mt}.

  3. Divide by the non-zero factor emte^{mt}

    9m2+6m+1=09m^{2}+6m+1=0

    emte^{mt} is never zero, so it can be cancelled.

  4. Note that the coefficients are copied straight across

    ah¨+bh˙+ch=0  am2+bm+c=0a\ddot{h}+b\dot{h}+ch=0\ \Rightarrow\ am^{2}+bm+c=0

    The order of each derivative becomes the power of mm.

  5. Read off the three coefficients

    a=9,b=6,c=1a=9,\quad b=6,\quad c=1

    They must be taken with their signs exactly as they appear in the model.

  6. Solve the auxiliary equation as a check

    m=13m=- \frac{1}{3}

    These roots are the exponents that appear in the complementary function.

  7. Check the discriminant is consistent with the model

    b24ac=0b^{2}-4ac=0

    Its sign classifies the damping, so it is a useful check on the coefficients.

  8. Reject any option with the wrong signs

    the signs must match the model exactly\text{the signs must match the model exactly}

    Changing a sign changes the roots and therefore the whole behaviour.

  9. Reject any option with the coefficients swapped

    am2+bm+ccm2+bm+aam^{2}+bm+c\neq cm^{2}+bm+a

    The leading coefficient comes from h¨\ddot{h} and the constant from hh; they cannot be interchanged.

  10. Reject any option that drops a term

    all three coefficients must appear\text{all three coefficients must appear}

    Losing the constant term would turn the quadratic into a different equation entirely.

  11. Select the auxiliary equation

    9m2+6m+1=09m^{2}+6m+1=0

    This is the quadratic whose roots give the exponents in the solution.

Answer
9m2+6m+1=09m^{2}+6m+1=0
Question 5
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

    b24ac=0b^{2}-4ac=0

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

  3. Write down the discriminant

    k24×9×4=k2144k^{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)e23tx=\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)e23t=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  b24ac=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  b24ac=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

    b24ac>0 heavy,b24ac=0 critical,b24ac<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

Unlock 65 more Modelling with differential equations questions

Create a free account to work through every Further Maths Modelling with differential equations question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Modelling with differential equations practice

Related Pure Maths topics