Modelling with differential equations Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Modelling with differential equations questions. See exactly how to solve problems on differential-equation-modelling, simple-harmonic-motion, period, frequency.

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.

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 marker buoy bobs vertically in still water. Its height hh metres above its mean position at time tt seconds satisfies h¨=−49h\ddot{h}=-49h. Find the exact period of the oscillation.

Worked solution

  1. Compare the model with the SHM equation

    h¨=−ω2hwithω2=49\ddot{h}=-\omega^{2}h\quad\text{with}\quad\omega^{2}=49

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

  2. Find the angular frequency

    ω=7\omega=7

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

  3. Use the period formula

    T=2πω=2π7T=\frac{2\pi}{\omega}=\frac{2\pi}{7}

    The period of SHM is fixed by ω\omega alone.

  4. State the period

    T=2π7T=\frac{2 \pi}{7}

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

Answer
T=2π7T=\frac{2 \pi}{7}
Question 3
2 markseasy
A long pendulum in a museum swings through a small angle. Its horizontal displacement xx metres from the vertical at time tt seconds satisfies x¨=−14x\ddot{x}=-\frac{1}{4}x. Find the exact period of the oscillation.

Worked solution

  1. Compare the model with the SHM equation

    x¨=−ω2xwithω2=14\ddot{x}=-\omega^{2}x\quad\text{with}\quad\omega^{2}=\frac{1}{4}

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

  2. Find the angular frequency

    ω=12\omega=\frac{1}{2}

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

  3. State the period

    T=4πT=4 \pi

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

Answer
T=4πT=4 \pi
Question 4
2 markseasy
A tuning fork prong vibrates. Its displacement xx millimetres from rest at time tt seconds satisfies x¨=−36x\ddot{x}=-36x. Find the frequency of the oscillation in hertz, giving your answer in exact form.

Worked solution

  1. Compare the model with the SHM equation

    x¨=−ω2xwithω2=36\ddot{x}=-\omega^{2}x\quad\text{with}\quad\omega^{2}=36

    The coefficient of xx is ω2\omega^{2}.

  2. Find the angular frequency

    ω=6\omega=6

    ω\omega is measured in radians per second.

  3. Convert the angular frequency to a frequency

    f=ω2π=62πf=\frac{\omega}{2\pi}=\frac{6}{2\pi}

    One complete oscillation corresponds to 2π2\pi radians of phase.

  4. State the frequency

    f=3πf=\frac{3}{\pi}

    This is the exact number of complete oscillations per second.

Answer
f=3πf=\frac{3}{\pi}
Question 5
2 markseasy
The bob of a metronome oscillates. Its angular displacement xx radians from the vertical at time tt seconds satisfies x¨=−4π2x\ddot{x}=-4 \pi^{2}x. Find the frequency of the oscillation in hertz, giving your answer in exact form.

Worked solution

  1. Compare the model with the SHM equation

    x¨=−ω2xwithω2=4π2\ddot{x}=-\omega^{2}x\quad\text{with}\quad\omega^{2}=4 \pi^{2}

    The coefficient of xx is ω2\omega^{2}.

  2. Find the angular frequency

    ω=2π\omega=2 \pi

    ω\omega is measured in radians per second.

  3. State the frequency

    f=1f=1

    This is the exact number of complete oscillations per second.

Answer
f=1f=1

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