Waves, Oscillations, Thermal Physics problem-solving question

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]

(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...

(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 $3.5 \times 10^{-3}\text{ m}$.

Part (a) Hint 1: Frequency is the number of cycles per second. Hint 2: Look for the relationship between frequency and angular frequency in the spec. Hint 3: $\omega = 2\pi f$. Part (b) Hint 1: Use the wave speed equation. Hint 2: For the explanation, think about the direction of particle vibration compared to the wave's travel. Hint 3: Progressive waves move energy; longitudinal waves vibrate "along" the line of travel. Part (c) Hint 1: The conversion factor between Celsius and Kelvin is $273.15$. Hint 2: Thermal equilibrium occurs when two objects are at the same temperature. Hint 3: If there is no net heat flow, the temperatures must have equalized. Part (d) Hint 1: The "new period"...