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Oscillations: Question 3

Syllabus 17.1, 17.2

Structured A2 9 marks

A simple pendulum consists of a small bob of mass m=0.120 kgm = 0.120\text{ kg} on a light string of length L=0.900 mL = 0.900\text{ m}. The bob is displaced sideways to give an amplitude of x0=5.0 cmx_0 = 5.0\text{ cm} and released from rest, so that it swings with simple harmonic motion. Take g=9.81 m s−2g = 9.81\text{ m s}^{-2}.

(a) Show that the period of oscillation is 1.90 s1.90\text{ s}. [2]

(b) Calculate the angular frequency ω\omega of the oscillation. [2]

(c) Calculate the total energy of the oscillation. [2]

(d) Calculate the kinetic energy and the potential energy of the bob at the instant its displacement from the equilibrium position is 3.0 cm3.0\text{ cm}. [3]

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Worked solution

Part (a): Show that T = 1.90 s

For a simple pendulum: T=2πLg=2π0.9009.81T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{0.900}{9.81}}

First find the ratio: 0.9009.81=0.0917 (3 s.f.)\dfrac{0.900}{9.81}=0.0917\ (3\text{ s.f.}). Taking the square root: 0.0917=0.3029\sqrt{0.0917}=0.3029.

T=2π×0.3029=1.903 sT=2\pi\times0.3029=1.903\text{ s}

Recompute as a check, splitting the square root differently: 0.900=0.9487\sqrt{0.900}=0.9487 and 9.81=3.1321\sqrt{9.81}=3.1321, so L/g=0.9487/3.1321=0.3029\sqrt{L/g}=0.9487/3.1321=0.3029, the same value. Multiplying by 2π2\pi again gives 1.903 s1.903\text{ s}.

So T=1.90 sT=1.90\text{ s} (3 s.f.), as required to show.

Part (b): Angular frequency

ω=gL=9.810.900=10.9=3.302 rad s−1\omega=\sqrt{\frac{g}{L}}=\sqrt{\frac{9.81}{0.900}}=\sqrt{10.9}=3.302\text{ rad s}^{-1}

Check using part (a): ω=2πT=2π1.903=3.302 rad s−1\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{1.903}=3.302\text{ rad s}^{-1}, the same value (small rounding only from using the rounded TT). So ω=3.30 rad s−1\omega=3.30\text{ rad s}^{-1} (3 s.f.).

Part (c): Total energy

The total (constant) energy of an SHM oscillation is: E=12mω2x02E=\tfrac12m\omega^2x_0^2

Substituting m=0.120 kgm=0.120\text{ kg}, ω=3.302 rad s−1\omega=3.302\text{ rad s}^{-1}, x0=0.050 mx_0=0.050\text{ m}: E=12×0.120×(3.302)2×(0.050)2=12×0.120×10.9×0.0025E=\tfrac12\times0.120\times(3.302)^2\times(0.050)^2=\tfrac12\times0.120\times10.9\times0.0025

Working step by step: 0.5×0.120=0.0600.5\times0.120=0.060; then 0.060×10.9=0.6540.060\times10.9=0.654; then 0.654×0.0025=1.635×10−3 J0.654\times0.0025=1.635\times10^{-3}\text{ J}.

Recompute in a different grouping as a check: (0.050)2×(3.302)2=0.0025×10.9=0.02725(0.050)^2\times(3.302)^2=0.0025\times10.9=0.02725; then 0.5×0.120×0.02725=0.06×0.02725=1.635×10−3 J0.5\times0.120\times0.02725=0.06\times0.02725=1.635\times10^{-3}\text{ J}, the same result.

So E=1.64×10−3 JE=1.64\times10^{-3}\text{ J} (3 s.f.), i.e. 1.64 mJ1.64\text{ mJ}.

Part (d): Kinetic and potential energy at x = 3.0 cm

Convert 3.0 cm=0.030 m3.0\text{ cm}=0.030\text{ m}.

Potential energy at displacement xx is PE=12mω2x2PE=\tfrac12m\omega^2x^2: PE=12×0.120×10.9×(0.030)2=0.654×0.0009=5.89×10−4 J (3 s.f.)PE=\tfrac12\times0.120\times10.9\times(0.030)^2=0.654\times0.0009=5.89\times10^{-4}\text{ J}\ (3\text{ s.f.})

Kinetic energy is the remainder of the total energy: KE=E−PE=1.635×10−3−0.589×10−3=1.05×10−3 J (3 s.f.)KE=E-PE=1.635\times10^{-3}-0.589\times10^{-3}=1.05\times10^{-3}\text{ J}\ (3\text{ s.f.})

Recompute KE independently as a check, using v=ωx02−x2v=\omega\sqrt{x_0^2-x^2} directly rather than subtraction: x02−x2=0.0025−0.0009=0.0016x_0^2-x^2=0.0025-0.0009=0.0016, so 0.0016=0.0400 m\sqrt{0.0016}=0.0400\text{ m}, giving v=3.302×0.0400=0.1321 m s−1v=3.302\times0.0400=0.1321\text{ m s}^{-1}. Then KE=12mv2=12×0.120×(0.1321)2=0.060×0.01745=1.05×10−3 JKE=\tfrac12mv^2=\tfrac12\times0.120\times(0.1321)^2=0.060\times0.01745=1.05\times10^{-3}\text{ J}, the same value as found by subtraction, confirming the result.

Final answers

  • (a) T=1.90 sT=\boxed{1.90}\text{ s}
  • (b) ω=3.30 rad s−1\omega=\boxed{3.30}\text{ rad s}^{-1}
  • (c) E=1.64×10−3 JE=\boxed{1.64\times10^{-3}}\text{ J}
  • (d) PE=5.89×10−4 JPE=\boxed{5.89\times10^{-4}}\text{ J}, KE=1.05×10−3 JKE=\boxed{1.05\times10^{-3}}\text{ J}