Marianopolis College · 203-SN3-RE

Marianopolis 203-SN3-RE: Waves and Modern Physics Tutoring

Why do stationary listeners at different distances from a stationary sound source measure different intensities but the same frequency? Questions like this require you to identify the quantity being measured before choosing an equation. For Marianopolis 203-SN3-RE, Nablio helps you connect the physical situation to a wave or particle model, then explain the calculation. The original sound example below separates power, intensity and decibel level.

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Connect the model to the measurement

Sound: distinguish source power, intensity at a position and the logarithmic decibel scale.
Light: explain which features of a problem call for wave behaviour and which call for photon energy.
Wave–particle duality: identify the experiment or observation that the chosen model must explain.
Nuclear and environmental applications: connect the quantities in your question to the relevant conservation or energy argument.

Check what changes when the distance changes

For an ideal point source radiating uniformly, the same power crosses spheres of increasing area. Intensity therefore falls as the inverse square of distance. Sound level is a logarithm of intensity, so it does not fall by the same numerical factor. State the model first: reflections, absorption and a directional source can change the result. Comparing measurements at two fixed positions does not by that change alone alter the frequency of a stationary source.

Worked example: sound level at a greater distance

An ideal stationary point source radiates sound uniformly in a reflection-free space. The measured sound level is 80.0 dB at 2.0 m. Find the intensity and sound level at 8.0 m. Neglect absorption and use the reference intensity $I_0=10^{-12}\,\mathrm{W/m^2}$ .

Let $I$ be sound intensity in $\mathrm{W/m^2}$ and $\beta$ sound level in decibels. Use subscript 1 for the measurement at 2.0 m and subscript 2 for the measurement at 8.0 m. Since $\beta=10\log_{10}(I/I_0)$ , the starting intensity is $I_1=I_0\,10^{80/10}=10^{-4}\,\mathrm{W/m^2}$ .
The distance becomes four times greater, so the area becomes sixteen times greater. Thus $I_2=I_1(2/8)^2=6.25\times10^{-6}\,\mathrm{W/m^2}$ .
Convert back to a sound level: $\beta_2=10\log_{10}(6.25\times10^6)\approx68.0\,\mathrm{dB}.$
Check with the distance ratio: $\beta_2-\beta_1=-20\log_{10}(8/2)\approx-12.0\,\mathrm{dB}$ . The source power and frequency have stayed the same; the intensity at the listener has decreased.

An intensity sixteen times smaller gives a level about 12 dB lower. Dividing 80 dB by sixteen would mix a logarithmic scale with a linear ratio.

Practise the ratio before using a calculator

Under the same assumptions, what are the intensity and sound level at 4.0 m? Would doubling the source power at that location restore the original 80.0 dB level?

Show the explained answer

At 4.0 m, the intensity is $2.5\times10^{-5}\,\mathrm{W/m^2}$ and the level is about 74.0 dB. Doubling the power doubles the intensity, adding about 3.01 dB to give 77.0 dB. Restoring 80.0 dB requires four times the original source power at this doubled distance.

Original Nablio example and practice, independent of the college. These are not course assessment questions. Follow your current section’s outline for the material to review.

Bring the graph, units and assumptions

Bring a question from your current outline and your attempted solution. We can locate the point where the physical model becomes unclear, check conversions between linear and logarithmic quantities, and practise explaining the result in words. This is independent practice, not a reproduction of a Marianopolis assessment.

We use permitted practice without completing graded assignments or exams on your behalf.

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