AS Waves 4.4.1(b)(i)4.4.1(b)(ii)4.4.1(c)4.4.1(d)4.4.1(g)

~12 min Difficulty: 5/10

Prior knowledge Basic wave definitions (amplitudewavelength)prefix meanings (micromilli).

Problem structure

(a) 3, numerical, frequency calculation from CRO. (b) 2, numerical, wave equation application. (c) 4, numerical, intensity-amplitude-distance relationship.

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A student uses a signal generator to produce a sound wave in a laboratory. The signal is monitored using a cathode-ray oscilloscope (CRO). The trace on the screen shows a clear sinusoidal wave. The time-base of the CRO is set to $250\text{ }\mu\text{s cm}^{-1}$ and the vertical gain is set to $5.0\text{ mV cm}^{-1}$. One full cycle of the wave on the screen occupies a horizontal distance of $3.2\text{ cm}$, and the peak-to-peak height of the trace is $4.8\text{ cm}$. (a) Calculate the frequency of the sound wave produced by the signal generator. [3] (b) The speed of sound in the laboratory is $344\text{ m s}^{-1}$. Determine the wavelength of this sound wave. [2] (c) The student moves a microphone, connected to the CRO, a distance of $0.50\text{ m}$ further away from the signal generator. The amplitude of the wave detected by the microphone decreases. Assuming the sound spreads out as a spherical wave from a point source, calculate the ratio: $$\frac{\text{Amplitude at new position}}{\text{Amplitude at original position}}$$ if the original distance from the source was $1.20\text{ m}$. [4]

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(a) First, determine the period $T$ from the CRO settings. The horizontal distance for one cycle is $3.2\text{ cm}$ and the time-base is $250\text{ }\mu\text{s cm}^{-1}$. $T = \text{horizontal distance} \times \text{time-base}$ $T = 3.2\text{ cm} \times 250 \times 10^{-6}\text{ s cm}^{-1} = 8.0 \times 10^{-4}\text{ s}$ Frequency $f = \frac{1}{T}$ $f = \frac{1}{8.0 \times 10^{-4}\text{ s}} = 1250\text{ Hz}$ Final answer: $1250\text{ Hz}$ (or $1.3\text{ kHz}$ to 2 SF). (b) Use the wave equation $v = f\lambda$. $\lambda = \frac{v}{f}$ $\lambda = \frac{344\text{ m s}^{-1}}{1250\text{ Hz}}$ $\lambda = 0.2752\text{ m}$ Final answer: $0.275\text{ m}$ (3 SF). (c) For a point source, Intensity $I \propto \frac{1}{r^2}$, where $r$ is the distance from the source. From the specification, Intensity $I \propto A^2$, where $A$ is amplitude. Therefore, $A^2 \propto \frac{1}{r^2}$, which implies $A \propto \frac{1}{r}$. Let $A_1$ be the amplitude at $r_1 = 1.20\text{ m}$. Let $A_2$ be the amplitude at $r_2 = 1.20 + 0.50 = 1.70\text{ m}$. The ratio is $\frac{A_2}{A_1} = \frac{1/r_2}{1/r_1} = \frac{r_1}{r_2}$ $\frac{A_2}{A_1} = \frac{1.20}{1.70} = 0.70588...$ Final answer: $0.71$ (2 SF).

Marking guidance

(a) M1: Correct calculation of period $T = 3.2 \times 250 \times 10^{-6}$ ($8.0 \times 10^{-4}\text{ s}$). M1: Use of $f = 1/T$. A1: Final answer $1250\text{ Hz}$ or $1.25\text{ kHz}$. (Allow 2 or 3 SF). Common error: Forgetting the micro ($\mu$) prefix or using the vertical height for frequency. (b) M1: Correct rearrangement or substitution into $v = f\lambda$. A1: $0.275\text{ m}$ (Allow ECF from part a). (c) M1: Recognition that Intensity $I \propto 1/r^2$. M1: Linking Intensity to Amplitude: $I \propto A^2$ (from spec). M1: Deducing $A \propto 1/r$ or setting up the ratio $\frac{A_2}{A_1} = \frac{r_1}{r_2}$. A1: Final ratio $0.71$ (Accept $0.706$). Common error: Using $I \propto 1/r$ or giving the ratio of intensities ($0.50$) instead of amplitudes.