University Physics III · Midterm Examination

A block on a light horizontal spring moves in simple harmonic motion on a frictionless surface. At one instant its speed is 60% of its maximum speed. What fraction of the oscillator's total mechanical energy is stored as elastic potential energy at that instant?

1. A0.36
2. B0.40
3. C0.80
4. D0.64

A lightly damped oscillator of mass m and natural angular frequency ω₀ is driven by a force F₀cos ωt with ω = 3ω₀. Which statement correctly describes its steady-state displacement?

1. AIts amplitude is about F₀/(8mω₀²), and it lags the driving force by nearly π.
2. BIts amplitude is about F₀/(8mω₀²), and it lags the driving force by π/2.
3. CIts amplitude is about F₀/(9mω₀²), and it lags the driving force by nearly π.
4. DIts amplitude is about F₀/(8mω₀²), and it is nearly in phase with the driving force.

A narrow tube is either open at both ends or closed at one end; you are not told which. Two of its successive resonant frequencies are 497 Hz and 639 Hz, with no resonance between them. Ignoring end corrections, what is the lowest resonant frequency of the air column?

1. A35.5 Hz
2. B71.0 Hz
3. C142 Hz
4. D355 Hz

Two small loudspeakers are driven in phase by the same signal generator at a single frequency. A point P is the same distance from both, and reflections are negligible. With either loudspeaker operating alone, the sound-intensity level at P is 70.0 dB. What is the level at P when both operate together?

1. A70.0 dB
2. B73.0 dB
3. C76.0 dB
4. D140 dB

Two loudspeakers emit steady tones of equal amplitude at 396.0 Hz and 401.0 Hz, and a microphone records their superposition. Which statement describes the recorded signal?

1. AA 398.5 Hz tone whose loudness peaks 2.5 times per second
2. BA 797 Hz tone whose loudness peaks 5 times per second
3. CA 398.5 Hz tone whose loudness peaks 5 times per second
4. DA steady 5.0 Hz tone, below the range of hearing

A step-index optical fibre has a core of refractive index 1.52 and a cladding of refractive index 1.46. Its flat entry face is perpendicular to its axis and is in air of refractive index 1.00. What is the largest angle to the fibre axis at which a ray in air can enter the core and still be guided by total internal reflection at the core–cladding boundary?

1. A16.2°
2. B41.1°
3. C73.8°
4. D25.0°

Unpolarized light of intensity I₀ passes in turn through three ideal linear polarizers whose transmission axes are at 30°, 90° and 135° to the vertical, all measured in the same rotational sense. What intensity emerges from the third polarizer?

1. AI₀/16
2. BI₀/8
3. CI₀/(4√2)
4. D0

An aluminium bar 1.50 m long fits snugly, and without stress, between two rigid supports at 12 °C. It is heated uniformly to 47 °C, and the supports prevent any change in its length. For aluminium, Young's modulus is 7.0 × 10¹⁰ Pa and the coefficient of linear expansion is 2.3 × 10⁻⁵ K⁻¹. What is the compressive stress in the bar?

1. A3.8 × 10⁷ Pa
2. B5.6 × 10⁷ Pa
3. C8.5 × 10⁷ Pa
4. D5.2 × 10⁸ Pa

A vibrating-wire strain gauge bolted to a concrete beam contains a steel wire of length L = 0.150 m and linear mass density μ = 3.85 × 10⁻⁴ kg/m, clamped at both ends. An electromagnet plucks the wire and a pick-up coil measures its fundamental frequency, which is 800 Hz when the gauge is installed.

1. (a)

   State State the linear wave equation obeyed by small transverse displacements y(x, t) of a wire under tension T with linear mass density μ, and the wave speed it implies.

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   1. 1

      ∂²y/∂t² = (T/μ) ∂²y/∂x², so the wave speed is v = √(T/μ)

2. (b)

   Calculate Calculate the speed of transverse waves on the wire and the tension in the wire when the gauge is installed.

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   1. 1

      both clamped ends are nodes, so the fundamental has λ₁ = 2L = 0.300 m and v = f₁λ₁ = 800 × 0.300 = 240 m/s

   2. 2

      T = μv² = 3.85 × 10⁻⁴ × 240² = 22.2 N

3. (c)

   Predict Loading the beam raises the tension in the wire by 3.0%. Treating L and μ as unchanged, predict the new fundamental frequency.

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   1. 1

      with L and μ fixed, f₁ = (1/2L)√(T/μ), so f₁ is proportional to √T

   2. 2

      f₁ = 800 × √1.030 = 812 Hz (811.9 Hz), a rise of about 1.5%

4. (d)

   Calculate In a separate test, a long wire with the same μ and the same tension as in (b) carries a sinusoidal transverse travelling wave of frequency 800 Hz and amplitude 0.20 mm, and its far end is terminated so that no wave is reflected. Calculate the average power the wave transmits along the wire.

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   1. 1

      ω = 2π × 800 = 5.03 × 10³ rad/s

   2. 2

      uses the average transmitted power P = ½μω²A²v

   3. 3

      P = ½ × 3.85 × 10⁻⁴ × (5.03 × 10³)² × (2.0 × 10⁻⁴)² × 240 = 4.7 × 10⁻² W (46.7 mW)

A cylindrical instrument buoy floats upright in calm sea water. Its mass is m = 850 kg, its horizontal cross-sectional area at the waterline is A = 0.500 m², and its sides are vertical from 0.40 m below to 0.40 m above the calm waterline. Take the density of sea water as ρ = 1025 kg/m³ and g = 9.81 m/s². Model the water as exerting a hydrostatic buoyant force together with a linear damping force of magnitude b|ż| that opposes the buoy's vertical velocity, ignore any water carried along with the buoy, and write γ = b/m and Q = ω₀/γ, where ω₀ is the undamped angular frequency.

1. (a)

   Derive Ignoring damping, derive from Newton's second law and Archimedes' principle the equation of motion for a small vertical displacement z of the buoy from its floating equilibrium, and hence identify ω₀.

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   1. 1

      at equilibrium the buoyant force ρgV₀ on the submerged volume V₀ balances the weight mg

   2. 2

      a further downward displacement z submerges an extra volume Az because the sides are vertical, so the net force is −ρgAz: a restoring force proportional to the displacement

   3. 3

      Newton's second law gives m d²z/dt² = −ρgAz, the simple-harmonic equation d²z/dt² = −ω₀²z with ω₀ = √(ρgA/m)

2. (b)

   Calculate Calculate the undamped period of the buoy's vertical oscillation.

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   1. 1

      ρgA = 1025 × 9.81 × 0.500 = 5.03 × 10³ N/m, so ω₀ = √(5.03 × 10³/850) = 2.43 rad/s

   2. 2

      T = 2π/ω₀ = 2.58 s

3. (c)

   Determine The buoy is pushed down 0.120 m in calm water and released. After 5 complete oscillations its amplitude has fallen to 0.030 m. Determine γ and Q.

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   1. 1

      for light damping the amplitude decays as z₀e^(−γt/2), and 5 oscillations take 5T = 12.9 s (the damped period differs from T by about 0.1%)

   2. 2

      γ = 2 ln(0.120/0.030)/(5T) = 2 × 1.386/12.92 = 0.215 s⁻¹

   3. 3

      Q = ω₀/γ = 2.43/0.215 = 11.3

4. (d)

   Predict A regular train of surface waves of amplitude η₀ = 0.050 m, with period equal to your answer to (b), now passes the buoy. Taking the driving force as ρgAη, where η = η₀cos ωt is the vertical displacement of the water surface, predict the amplitude of the buoy's steady-state oscillation.

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   1. 1

      at resonance the steady-state amplitude is F₀/(bω₀) = F₀/(mγω₀), with F₀ = ρgAη₀ = 5.03 × 10³ × 0.050 = 251 N

   2. 2

      since ρgA = mω₀², this is Qη₀ = 11.3 × 0.050 = 0.57 m

5. (e)

   Evaluate Evaluate whether the model used in (d) can be trusted for these waves.

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   1. 1

      the predicted 0.57 m amplitude exceeds the 0.40 m over which the sides are vertical (relative to the moving surface the excursion is also 0.57 m, since z lags η by π/2 at resonance and √(0.57² + 0.050²) = 0.57 m), so the waterline area, and with it the restoring force per unit displacement, would change during each cycle: the linear restoring force the model rests on breaks down

   2. 2

      judgement: the 0.57 m prediction is not reliable; at resonance the linear model holds only for wave amplitudes below about 0.40 m/Q ≈ 0.035 m

A discharge lamp contains a mixture of hydrogen and deuterium. Its blue-green Balmer emission consists of two lines, at λ₁ = 486.13 nm from hydrogen and λ₂ = 486.00 nm from deuterium. To show that both lines are present, a student may use either a transmission grating ruled with 1200 lines per millimetre over a width of 25.0 mm, used at normal incidence and able to be illuminated over any chosen width, or a Michelson interferometer whose movable mirror can be moved through up to 4.0 mm from the position at which the two arms are equal in length.

1. (a)

   Calculate Calculate the chromatic resolving power needed to separate the two lines.

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   1. 1

      R = λ/Δλ, with λ ≈ 486.07 nm and Δλ = 486.13 − 486.00 = 0.13 nm

   2. 2

      R = 486.07/0.13 = 3.7 × 10³ (3739)

2. (b)

   Determine Determine the highest diffraction order the grating produces at this wavelength, and hence the minimum width of grating that must be illuminated to resolve the lines.

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   1. 1

      d = 1/(1200 mm⁻¹) = 833 nm, and at normal incidence sin θ = mλ/d ≤ 1 requires m ≤ d/λ = 1.71, so only the first order exists

   2. 2

      R = mN with m = 1 requires at least N = 3739 illuminated lines

   3. 3

      minimum width = Nd = 3739 × 833 nm = 3.12 mm, well within the 25.0 mm available

3. (c)

   Show that As the mirror moves a distance x from the position of equal arm lengths, the detector records fringes whose visibility falls periodically to a minimum and recovers. Show that the mirror displacement between successive visibility minima is λ₁λ₂/[2(λ₁ − λ₂)] ≈ λ²/(2Δλ).

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   1. 1

      moving the mirror by x changes the path difference by 2x, so the interference orders of the two lines at the detector are 2x/λ₁ and 2x/λ₂

   2. 2

      the two fringe systems are in step (maximum visibility) when the orders differ by a whole number and exactly out of step (minimum visibility) when they differ by a half-integer; the difference is 2x/λ₂ − 2x/λ₁ = 2x(λ₁ − λ₂)/(λ₁λ₂)

   3. 3

      successive minima are reached each time this difference grows by 1, so their spacing is Δx = λ₁λ₂/[2(λ₁ − λ₂)] ≈ λ²/(2Δλ)

4. (d)

   Sketch Assuming the two lines are of equal intensity, sketch a graph of fringe visibility against mirror displacement from 0 to 4.0 mm, marking the displacements at which the visibility is a minimum.

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   1. 1

      visibility is 1 at zero displacement and follows |cos(2πxΔλ/λ²)|: rounded maxima, falling to a sharp zero at 0.45 mm and recovering to 1 at 0.91 mm

   2. 2

      further zeros marked at 1.36, 2.27 and 3.18 mm (spacing 0.91 mm), with maxima of 1 at 1.82, 2.73 and 3.63 mm: four minima within the 4.0 mm travel

Reflective road markings contain transparent glass beads, each a sphere of radius R = 0.300 mm, half-embedded in white paint. Light from a car's headlamps enters the exposed front of a bead and is focused towards the back surface, where the paint scatters it; light scattered from a small bright patch there is sent back towards the car by the bead. Treat the incident light as a parallel beam travelling along an axis through the bead's centre, in air of refractive index 1.00. For refraction at a single spherical surface use n₁/s + n₂/s′ = (n₂ − n₁)/R, where s is positive for a real object in front of the surface, s′ is positive for a real image behind it, and R is positive when the centre of curvature lies behind the surface.

1. (a)

   Show that Show that, in the paraxial limit, refraction at the front surface alone would bring the beam to a focus a distance nR/(n − 1) behind that surface, where n is the refractive index of the glass.

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   1. 1

      for a parallel beam s → ∞, so n₁/s = 0, with n₁ = 1.00, n₂ = n, and R positive because the centre of curvature of the front surface lies inside the bead, behind that surface

   2. 2

      n/s′ = (n − 1)/R, so s′ = nR/(n − 1)

2. (b)

   Determine Determine the refractive index a bead must have for the paraxial focus to lie exactly on the paint at its back surface, and state where the paraxial focus lies for a bead of ordinary glass with n = 1.50.

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   1. 1

      the back surface is a full diameter, 2R, behind the front surface, so n/(n − 1) = 2, giving n = 2.00

   2. 2

      for n = 1.50, s′ = 3R = 0.900 mm, which is 0.300 mm beyond the back surface, inside the paint, so the illuminated patch on the paint is broad

3. (c)

   Show that Take the centre of the bead as origin, with the incident light travelling in the +x direction, so that the axis meets the back surface at (R, 0). A ray entering parallel to the axis at height h above it has angle of incidence i, where sin i = h/R, and angle of refraction r. Without making the paraxial approximation, show that this ray reaches the back surface at the point whose radius makes an angle i − 2r with the +x axis, on the opposite side of the axis from the point of entry.

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   1. 1

      the normal at the entry point lies along the radius, so that radius makes an angle i with the −x axis, i.e. an angle 180° − i measured from the +x axis on the side of the entry point

   2. 2

      the path inside the bead is a chord, which forms an isosceles triangle with the radii to its two ends; both base angles equal r, so the angle at the centre between the entry and exit radii is 180° − 2r

   3. 3

      the ray is bent towards the axis, so the exit radius lies 180° − 2r further round, at (180° − i) − (180° − 2r) = −(i − 2r) from the +x axis: an angle i − 2r on the opposite side of the axis from the entry point

4. (d)

   Calculate For a bead with n = 2.00, calculate the perpendicular distance from the axis of the point at which the back surface is reached by the ray entering at h = 0.50R, and by the ray entering at h = 0.90R.

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   1. 1

      h = 0.50R: i = 30.0°, r = sin⁻¹(0.25) = 14.48°, i − 2r = 1.04°, so the distance is R sin(i − 2r) = 0.300 × 0.0182 = 5.5 × 10⁻³ mm (5.5 μm) beyond the axis

   2. 2

      h = 0.90R: i = 64.16°, r = sin⁻¹(0.45) = 26.74°, i − 2r = 10.67°, so the distance is 0.300 × sin 10.67° = 5.6 × 10⁻² mm (56 μm) beyond the axis

5. (e)

   Evaluate High-index road-marking beads are typically made from glass with n close to 1.9 rather than exactly 2.00. Evaluate this choice in the light of your results.

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   1. 1

      at n = 2.00 only near-axis rays reach the axis point; rays entering further out cross the axis before the back surface and land up to 56 μm from it at h = 0.90R, so satisfying the paraxial condition exactly does not give the smallest patch on the paint

   2. 2

      lowering n to 1.90 moves the paraxial focus just beyond the back surface (s′ = 2.11R), so near-axis rays land slightly on the entry side of the axis while outer rays land less far on the opposite side: at n = 1.90 the ray entering at h = 0.90R has r = 28.27° and i − 2r = 7.61°, landing 0.300 × sin 7.61° = 40 μm beyond the axis against 56 μm at n = 2.00, so the patch is narrower (about 44 μm across for h ≤ 0.90R), returning a tighter, brighter cone to the driver, and the choice is sound

To measure the speed of sound outdoors, a group mounts a buzzer of steady but unknown frequency f₀ on the roof of a car. On a calm day at 21 °C the car is driven at a series of steady speeds vₛ along a straight runway directly towards a microphone at the same height, and a spectrum analyser records the frequency f received while the car is still far away. Each speed is logged by satellite navigation. In still air the moving-source Doppler relation is f = f₀v/(v − vₛ), where v is the speed of sound.

1. (a)

   Show that Show that the relation predicts a straight line when 1/f is plotted against vₛ, and identify its gradient and its intercept on the 1/f axis.

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   1. 1

      inverting, 1/f = (v − vₛ)/(f₀v) = 1/f₀ − vₛ/(f₀v), which is linear in vₛ

   2. 2

      gradient = −1/(f₀v); intercept = 1/f₀

2. (b)

   Determine Use the readings to determine f₀ and the speed of sound v.

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   1. 1

      a best-fit straight line of 1/f against vₛ, with its gradient read from widely separated points on the line, gives gradient = −1.35 × 10⁻⁶ s² m⁻¹ (accept −1.34 × 10⁻⁶ to −1.36 × 10⁻⁶ s² m⁻¹)

   2. 2

      the intercept at vₛ = 0 is 1/f₀ = 4.651 × 10⁻⁴ s, so f₀ = 2150 Hz (accept 2148 Hz to 2152 Hz)

   3. 3

      v = −intercept/gradient = 1/(f₀ × |gradient|)

   4. 4

      v = 4.651 × 10⁻⁴/1.35 × 10⁻⁶ = 344 m/s (accept 341 m/s to 348 m/s), consistent with the 344 m/s expected in air at 21 °C

3. (c)

   Justify Justify, using the readings, the use of the exact Doppler relation rather than the low-speed approximation f ≈ f₀(1 + vₛ/v).

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   1. 1

      the approximation makes f linear in vₛ, rising by the same f₀ × 4.0/v ≈ 25.0 Hz for each 4.0 m/s step and reaching 2300.0 Hz at 24.0 m/s; the readings rise by 25.4, 26.8, 26.8, 28.3 and 28.3 Hz and reach 2311.2 Hz

   2. 2

      the steadily growing steps and the 11 Hz excess at 24.0 m/s are many times the scatter of the readings about the reciprocal straight line (about 0.3 Hz), so the data reject the approximation, which drops terms in (vₛ/v)² and higher, and the exact relation is needed

4. (d)

   Predict On a second day a steady wind of 5.0 m/s blows along the runway from the microphone towards the approaching car. Predict how the intercept and the gradient of the plot of 1/f against vₛ would change, and the value of v that the analysis in (b) would then return.

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   1. 1

      in the frame of the air the microphone moves away from the car at 5.0 m/s and the car approaches at vₛ + 5.0 m/s, so f = f₀(v − 5.0)/(v − 5.0 − vₛ); at vₛ = 0 the buzzer and microphone share the same motion through the air, f = f₀, and the intercept is unchanged

   2. 2

      1/f = 1/f₀ − vₛ/[f₀(v − 5.0)], so the gradient is steeper and the analysis returns v − 5.0 = 339 m/s: the speed of sound relative to the ground along the path, not the speed of sound in the air

A sealed glass bulb of volume V = 1.50 × 10⁻³ m³ contains argon at a temperature of 293 K and an initial pressure of 3.0 Pa. A circular hole 60 μm in diameter is then opened in a thin foil in the bulb's wall, leading directly to a high vacuum, while the bulb is held at 293 K. Model the escape as effusion, in which atoms leave through the hole independently of one another, and treat the argon as an ideal gas with a Maxwell–Boltzmann distribution of speeds, whose mean speed is ⟨v⟩ = √(8kT/(πm)). Take the Boltzmann constant k = 1.381 × 10⁻²³ J/K, the mass of an argon atom as 39.95 u with 1 u = 1.661 × 10⁻²⁷ kg, and the effective diameter of an argon atom as about 0.36 nm.

1. (a)

   Show that Consider a gas with n atoms per unit volume and an isotropic distribution of velocities. Show that the number of atoms striking unit area of the wall per unit time is Φ = ¼n⟨v⟩.

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   1. 1

      isotropy: the fraction of atoms moving at between θ and θ + dθ to the inward normal of the wall is 2π sin θ dθ/4π = ½ sin θ dθ

   2. 2

      atoms with speeds between v and v + dv moving in that direction reach an area A of wall within a time dt only if they lie in a slanted cylinder of volume Av cos θ dt, so they make n g(v)dv × ½ sin θ dθ × Av cos θ dt impacts, where g(v) is the normalised speed distribution

   3. 3

      integrating over the directions heading towards the wall, ∫₀^(π/2) ½ sin θ cos θ dθ = ¼

   4. 4

      integrating over speeds, ∫₀^∞ v g(v) dv = ⟨v⟩, so Φ = ¼n⟨v⟩

2. (b)

   Derive Derive, from conservation of the number of atoms and the ideal-gas equation of state, an expression for the pressure in the bulb as a function of time t, identifying the time constant τ in terms of V, the area A of the hole and ⟨v⟩.

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   1. 1

      atoms leave through the hole at the rate ΦA and none return from the vacuum, so dN/dt = −¼(N/V)⟨v⟩A

   2. 2

      at constant T, ⟨v⟩ is constant, so integrating gives N = N₀e^(−t/τ) with τ = 4V/(A⟨v⟩)

   3. 3

      pV = NkT at constant V and T makes p proportional to N, so p = p₀e^(−t/τ)

3. (c)

   Calculate Calculate ⟨v⟩ for the argon, the time constant τ, and the time taken for the pressure to fall from 3.0 Pa to 0.30 Pa.

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   1. 1

      m = 39.95 × 1.661 × 10⁻²⁷ = 6.636 × 10⁻²⁶ kg, so ⟨v⟩ = √(8 × 1.381 × 10⁻²³ × 293/(π × 6.636 × 10⁻²⁶)) = 394 m/s

   2. 2

      A = π(30 × 10⁻⁶)² = 2.83 × 10⁻⁹ m², so τ = 4 × 1.50 × 10⁻³/(2.83 × 10⁻⁹ × 394) = 5.39 × 10³ s (about 1.5 h)

   3. 3

      t = τ ln(3.0/0.30) = 5.39 × 10³ × 2.303 = 1.24 × 10⁴ s (about 3.4 h)

4. (d)

   Estimate Estimate the mean free path of the argon atoms at the initial pressure, stating the model you use, and hence decide whether treating the escape as effusion is justified.

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   1. 1

      model: hard spheres of diameter d ≈ 0.36 nm with Maxwell–Boltzmann relative speeds, so λ = kT/(√2πd²p), with number density p/kT = 7.4 × 10²⁰ m⁻³

   2. 2

      λ = 1.381 × 10⁻²³ × 293/(√2 × π × (0.36 × 10⁻⁹)² × 3.0) ≈ 2.3 × 10⁻³ m (about 2 mm; accept about 3 mm if the √2 is omitted)

   3. 3

      λ is about 40 times the 60 μm diameter of the hole, so atoms pass through it without colliding with one another; the effusion model is justified, and becomes more so as the pressure falls

5. (e)

   Explain If the bulb were thermally insulated instead of held at 293 K, the argon remaining in it would cool as it leaked. Explain why, given that the gas does no work on any piston.

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   1. 1

      the flux weights atoms by their speed, since the rate at which atoms of speed v arrive at the hole is proportional to v g(v); faster atoms escape preferentially, and each escaping atom carries a mean kinetic energy of 2kT, more than the average of 3kT/2 per atom in the bulb

   2. 2

      the gas left behind therefore loses more than the average energy per atom removed, so its mean translational kinetic energy per atom, (3/2)kT, falls and its temperature falls with it