Move away from a sound source, from distance r, to r₂ = 2₁. (A) By what factor does the intensity change? (B) What is the change in sound level in decibels? But suppose the distance increases by the factor 4.3. (A) Now what is the change in sound intensity? (B) And what is the change in sound level, in decibels?
Properties of sound
A sound wave is a mechanical wave (or mechanical vibration) that transit through media such as gas (air), liquid (water), and solid (wood).
Quality Of Sound
A sound or a sound wave is defined as the energy produced due to the vibrations of particles in a medium. When any medium produces a disturbance or vibrations, it causes a movement in the air particles which produces sound waves. Molecules in the air vibrate about a certain average position and create compressions and rarefactions. This is called pitch which is defined as the frequency of sound. The frequency is defined as the number of oscillations in pressure per second.
Categories of Sound Wave
People perceive sound in different ways, like a medico student takes sound as vibration produced by objects reaching the human eardrum. A physicist perceives sound as vibration produced by an object, which produces disturbances in nearby air molecules that travel further. Both of them describe it as vibration generated by an object, the difference is one talks about how it is received and other deals with how it travels and propagates across various mediums.
![**Sound Intensity and Level Changes with Distance**
When moving away from a sound source, the distance from the source changes from \( r_1 \) to \( r_2 = 2r_1 \).
**(A) By what factor does the intensity change?**
Sound intensity is inversely proportional to the square of the distance from the source. Therefore, as the distance doubles, the intensity decreases by a factor of \((2^2) = 4\).
**(B) What is the change in sound level in decibels?**
The change in sound level in decibels is given by the formula:
\[ \Delta L = 10 \log_{10}\left(\frac{I_2}{I_1}\right) \]
where \( I_2/I_1 = 1/4 \) since the intensity decreases by a factor of 4.
Substituting into the formula:
\[ \Delta L = 10 \log_{10}\left(\frac{1}{4}\right) \approx -6 \text{ dB} \]
This means the sound level decreases by approximately 6 decibels.
**Scenario with a Different Factor:**
Suppose the distance increases by the factor 4.3.
**(A) Now what is the change in sound intensity?**
The intensity decreases by a factor of \((4.3^2) \approx 18.49\).
**(B) And what is the change in sound level, in decibels?**
Using the same formula for decibels:
\[ \Delta L = 10 \log_{10}\left(\frac{1}{18.49}\right) \approx -12.6 \text{ dB} \]
Thus, the sound level decreases by approximately 12.6 decibels.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F29d54699-8096-4754-af5c-5369115407dd%2F33323545-f223-46bd-90ae-cdce010c4ce9%2Fpl7tro_processed.png&w=3840&q=75)
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