A2 WavesOscillationsThermal Physics 5.3.1(a)5.3.1(b)5.3.1(c)(i)4.4.1(a)4.4.1(b)(i)

~20 min Difficulty: 6/10 13 marks

Prior knowledge Basic wave properties and $F=ma$.

Problem structure

(a) 2, numerical, angular frequency calculation. (b)(i) 2, numerical, wave equation application. (b)(ii) 2, qualitative, wave definitions. (c)(i) 2, numerical, temperature scale conversion. (c)(ii) 2, qualitative/numerical, thermal equilibrium. (d) 3, numerical, SHM acceleration calculation.

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A specialized acoustic sensor consists of a small metal sphere of mass $m = 1.20 \times 10^{-2}\text{ kg}$ suspended by a spring. When the sphere is displaced and released, it undergoes simple harmonic motion (SHM). An engineer uses an oscilloscope to monitor the displacement of the sphere over time. The oscilloscope trace shows that the sphere completes $5.0$ full oscillations in a time interval of $0.40\text{ s}$. (a) Calculate the angular frequency $\omega$ of the sphere's oscillation. [2] (b) The sensor is placed in a chamber where it is bombarded by a longitudinal sound wave. The sound wave has a wavelength of $0.68\text{ m}$ and travels at a speed of $340\text{ m s}^{-1}$. (i) Determine the frequency of this sound wave. [2] (ii) Explain, with reference to the motion of particles in the medium, why this sound wave is classified as a longitudinal progressive wave. [2] (c) The sensor is initially at a room temperature of $22.0\text{ }^\circ\text{C}$. It is then placed into a thermal bath at a temperature of $315.15\text{ K}$. (i) Convert the room temperature into kelvin and the thermal bath temperature into degrees Celsius. [2] (ii) The sensor and the bath are left until they reach thermal equilibrium. State what is meant by thermal equilibrium in this context and identify the final temperature of the sensor in kelvin. [2] (d) Due to thermal expansion at the higher temperature, the spring constant changes, causing the new period of the sphere's SHM to become exactly equal to the period of the sound wave described in part (b). Calculate the required acceleration of the sphere when its displacement from equilibrium is $3.5\text{ mm}$ at this new temperature. [3]

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A specialized acoustic sensor consists of a small metal sphere of mass $m = 1.20 \times 10^{-2}\text{ kg}$ suspended by a spring. When the sphere is displaced and released, it undergoes simple harmonic motion (SHM). An engineer uses an oscilloscope to monitor the displacement of the sphere over time. The oscilloscope trace shows that the sphere completes $5.0$ full oscillations in a time interval of $0.40\text{ s}$.

(a) Calculate the angular frequency $\omega$ of the sphere's oscillation.

[2]

(b) The sensor is placed in a chamber where it is bombarded by a longitudinal sound wave. The sound wave has a wavelength of $0.68\text{ m}$ and travels at a speed of $340\text{ m s}^{-1}$. (i) Determine the frequency of this sound wave. [2] (ii) Explain, with reference to the motion of particles in the medium, why this sound wave is classified as a longitudinal progressive wave.

[2]

(c) The sensor is initially at a room temperature of $22.0\text{ }^\circ\text{C}$. It is then placed into a thermal bath at a temperature of $315.15\text{ K}$. (i) Convert the room temperature into kelvin and the thermal bath temperature into degrees Celsius. [2] (ii) The sensor and the bath are left until they reach thermal equilibrium. State what is meant by thermal equilibrium in this context and identify the final temperature of the sensor in kelvin.

[2]

(d) Due to thermal expansion at the higher temperature, the spring constant changes, causing the new period of the sphere's SHM to become exactly equal to the period of the sound wave described in part (b). Calculate the required acceleration of the sphere when its displacement from equilibrium is $3.5\text{ mm}$ at this new temperature.

[3]

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(a) First, find the frequency $f$ of the oscillations: $f = \frac{\text{number of oscillations}}{\text{time}} = \frac{5.0}{0.40} = 12.5\text{ Hz}$ Using the relationship $\omega = 2\pi f$: $\omega = 2 \times \pi \times 12.5 = 78.539... \text{ rad s}^{-1}$ $\omega \approx 78.5\text{ rad s}^{-1}$ (3 SF) (b) (i) Using the wave equation $v = f\lambda$: $f = \frac{v}{\lambda} = \frac{340}{0.68} = 500\text{ Hz}$ (ii) It is a progressive wave because it transfers energy from one point to another without transferring the medium itself. It is longitudinal because the particles of the medium oscillate parallel to the direction of energy transfer (propagation). (c) (i) Using $T(\text{K}) = \theta(^\circ\text{C}) + 273.15$: Room temperature: $22.0 + 273.15 = 295.15\text{ K}$ Bath temperature: $315.15 - 273.15 = 42.0\text{ }^\circ\text{C}$ (ii) Thermal equilibrium means there is no net flow of thermal energy between the sensor and the bath. The final temperature of the sensor will be the same as the bath: $315.15\text{ K}$. (d) The new period $T_{new}$ is equal to the period of the sound wave. $T_{new} = \frac{1}{f_{sound}} = \frac{1}{500} = 2.00 \times 10^{-3}\text{ s}$ Calculate the new angular frequency $\omega_{new}$: $\omega_{new} = \frac{2\pi}{T_{new}} = 2\pi \times 500 = 3141.59... \text{ rad s}^{-1}$ Using the SHM definition $a = -\omega^2 x$: Magnitude of acceleration $|a| = \omega_{new}^2 x$ $|a| = (3141.59...)^2 \times (3.5 \times 10^{-3})$ $|a| = 9.8696... \times 10^6 \times 3.5 \times 10^{-3} = 34543.6... \text{ m s}^{-2}$ $a \approx 3.5 \times 10^4 \text{ m s}^{-2}$ (2 SF, matching the $3.5\text{ mm}$ precision).

Marking guidance

(a) 1 mark: Correct calculation of frequency ($12.5\text{ Hz}$) or period ($0.08\text{ s}$). 1 mark: Correct calculation of $\omega$ to at least 2 SF ($78.5$ or $79\text{ rad s}^{-1}$). (b) (i) 1 mark: Correct rearrangement of $v=f\lambda$. 1 mark: Final answer $500\text{ Hz}$. (ii) 1 mark: Definition of progressive (energy transfer). 1 mark: Definition of longitudinal (oscillation parallel to propagation). (c) (i) 1 mark: $295.15\text{ K}$ (allow $295\text{ K}$). 1 mark: $42.0\text{ }^\circ\text{C}$. (ii) 1 mark: Definition of thermal equilibrium (no net heat flow). 1 mark: $315.15\text{ K}$. (d) 1 mark: Identifying $f_{new} = 500\text{ Hz}$ or $T_{new} = 0.002\text{ s}$. 1 mark: Correct substitution into $a = -\omega^2 x$ (ignore sign for magnitude). 1 mark: Final answer $3.5 \times 10^4 \text{ m s}^{-2}$ (allow $34500$ to $34600$). Common error: Forgetting to convert $3.5\text{ mm}$ to $3.5 \times 10^{-3}\text{ m}$.