Waves problem-solving question

A student uses a signal generator to produce a sound wave in a laboratory. The wave is detected by a microphone connected to an oscilloscope. The oscilloscope screen shows a trace of the wave with the time-base set to $50\text{ \mu s cm}^{-1}$ and the vertical sensitivity set to $20\text{ mV cm}^{-1}$. The trace has a peak-to-peak height of $4.8\text{ cm}$ and the horizontal distance between three consecutive peaks is $13.6\text{ cm}$. (a) Determine the frequency of the sound wave and the amplitude of the signal in mV. [3] (b) The sound wave travels through air at a speed of $343\text{ m s}^{-1}$. Calculate the wavelength of the wave and the phase difference, in radians, between two points in the air separated by a distance of $12.5\text{ cm}$ along the direction of wave travel. [4] (c) The sound wave is incident normally on a rectangular aperture of area $0.025\text{ m}^2$. The total power of the wave passing through this area is $1.8 \times 10^{-5}\text{ W}$. If the amplitude of the wave is doubled while the frequency remains constant, calculate the new intensity of the wave. [3]

(a) The horizontal distance for two full cycles (three consecutive peaks) is $13.6\text{ cm}$. Therefore, the distance for one period $T$ on the screen is $d = 13.6 / 2 = 6.8\text{ cm}$. Using the time-base $B = 50\text{ \mu s cm}^{-1}$: $T = d \times B = 6.8\text{ cm} \times 50 \times 10^{-6}\text{ s cm}^{-1} = 3.4 \times 10^{-4}\text{ s}$. The frequency $f$ is: $f = \frac{1}{T} = \frac{1}{3.4 \times 10^{-4}} \approx 2941.17\text{ Hz}$. The peak-to-peak height is $4.8\text{ cm}$. The amplitude $A$ on the screen is half of this: $2.4\text{ cm}$. Using the vertical sensitivity $S = 20\text{ mV cm}^{-1}$: $V_0 = 2.4\text{ cm} \times 20\text{ mV cm}^{-1} = 48\text{ mV}$. Final values: $f = 2.9\text{ kHz}$ (or $2940\text{ Hz}$), $A = 48\text{ mV}$. (b) Using the wave equation $v = f\lambda$: $\lambda = \frac{v}{f} = \frac{343}{2941.17} \approx 0.11662\text{ m} = 11.66\text{ cm}$. The phase difference $\phi$ is given by the ratio of the separation $\Delta x$ to the wavelength $\lambda$, multiplied by $2\pi$: $\phi = \frac{\Delta x}{\lambda} \times 2\pi$ $\phi = \frac{12.5\text{ cm}}{11.66\text{ cm}} \times 2\pi \approx 1.0718 \times 2\pi = 6.734\text{ rad}$. To express as a principal phase difference (optional but good practice): $6.734 - 2\pi \approx 0.45\text{ rad}$. However, $6.7\text{ rad}$ is the direct physical phase shift. Final values: $\lambda = 0.117\text{ m}$, $\phi = 6.7\text{ rad}$. (c) Initial intensity $I_1 = \frac{P}{A} = \frac{1.8 \times 10^{-5}\text{ W}}{0.025\text{ m}^2} = 7.2 \times 10^{-4}\text{ W m}^{-2}$. The relationship between intensity and amplitude...

(a) - 1 mark for calculating the period $T$ correctly from the trace ($3.4 \times 10^{-4}\text{ s}$). - 1 mark for $f = 1/T$ resulting in $2900\text{ Hz}$ or $2.9\text{ kHz}$ (allow ecf from $T$). - 1 mark for amplitude $V_0 = 48\text{ mV}$ (must be half of peak-to-peak). (b) - 1 mark for $\lambda = v/f$ resulting in $0.117\text{ m}$ (allow ecf from $f$). - 1 mark for the phase difference formula $\phi = 2\pi \Delta x / \lambda$. - 1 mark for correct substitution of $\Delta x$ and $\lambda$ in consistent units. - 1 mark for final answer $6.7\text{ rad}$ (accept $0.45\text{ rad}$ if reduced by $2\pi$). (c) - 1 mark for calculating initial intensity $I = 7.2 \times 10^{-4}\text{ W m}^{-2}$. - 1 mark for stating or using $I \propto A^2$. - 1 mark for final answer $2.9 \times 10^{-3}\text{ W m}^{-2}$ (allow ecf from initial $I$). Common errors: Using 3 peaks as 3 wavelengths (it is 2); forgetting to halve peak-to-peak for amplitude; using degrees instead of radians; forgetting the square...

(a) Hint 1: Look at the horizontal distance. If there are three peaks, how many full wave cycles are between the first and the third? (a) Hint 2: The time-base tells you how much time each cm on the screen represents. Use this to find the time for one cycle (the period). (a) Hint 3: Amplitude is the maximum displacement from the equilibrium, which is half of the total vertical "swing" (peak-to-peak). (b) Hint 1: Use the wave equation $v = f\lambda$ with the frequency you found in part (a). (b) Hint 2: Phase difference is a measure of how far "out of step" two points are. It is the fraction of a wavelength multiplied by $2\pi$. (b) Hint 3: Ensure your distance $\Delta x$ and wavelength...