Math Problem Statement

The intensity of the sound of a certain television advertisement is 10 times greater than the intensity of the television program. By how many decibels does the loudness increase? Use the function $L\left(I\right)=10\log\frac{I}{I_0}$ . The loudness increases by decibels.

Solution

We are given that the intensity of the sound of the television advertisement is 10 times greater than the intensity of the television program. We need to calculate the increase in decibels using the given loudness function:

L(I)=10log(II0)L(I) = 10 \log \left( \frac{I}{I_0} \right)

Where:

  • II is the intensity of the sound.
  • I0I_0 is the reference intensity (a constant).

Step-by-Step Solution:

Let IadI_{\text{ad}} represent the intensity of the advertisement sound, and IprogI_{\text{prog}} represent the intensity of the program sound. From the problem, we know:

Iad=10×IprogI_{\text{ad}} = 10 \times I_{\text{prog}}

We want to find the difference in loudness ΔL\Delta L between the advertisement and the program.

The loudness of the advertisement and the program can be written as:

Lad=10log(IadI0)L_{\text{ad}} = 10 \log \left( \frac{I_{\text{ad}}}{I_0} \right)

Lprog=10log(IprogI0)L_{\text{prog}} = 10 \log \left( \frac{I_{\text{prog}}}{I_0} \right)

Step 1: Find the difference in loudness

The difference in loudness ΔL\Delta L is given by:

ΔL=LadLprog\Delta L = L_{\text{ad}} - L_{\text{prog}}

Substitute the expressions for LadL_{\text{ad}} and LprogL_{\text{prog}}:

ΔL=10log(IadI0)10log(IprogI0)\Delta L = 10 \log \left( \frac{I_{\text{ad}}}{I_0} \right) - 10 \log \left( \frac{I_{\text{prog}}}{I_0} \right)

Use the logarithmic property logalogb=log(ab)\log a - \log b = \log \left( \frac{a}{b} \right):

ΔL=10log(IadIprog)\Delta L = 10 \log \left( \frac{I_{\text{ad}}}{I_{\text{prog}}} \right)

Step 2: Substitute Iad=10IprogI_{\text{ad}} = 10 I_{\text{prog}}

Substitute Iad=10IprogI_{\text{ad}} = 10 I_{\text{prog}} into the equation:

ΔL=10log(10IprogIprog)\Delta L = 10 \log \left( \frac{10 I_{\text{prog}}}{I_{\text{prog}}} \right)

Simplify the fraction:

ΔL=10log(10)\Delta L = 10 \log (10)

Step 3: Calculate the value of ΔL\Delta L

Since log(10)=1\log (10) = 1, we get:

ΔL=10×1=10 decibels\Delta L = 10 \times 1 = 10 \text{ decibels}

Final Answer:

The loudness increases by 10 decibels.


Would you like further explanation or more details on any step?

Here are 5 related questions you can explore:

  1. How does the loudness change if the intensity is 100 times greater instead of 10 times?
  2. What happens to the loudness if the intensity decreases instead of increases?
  3. How would you calculate the loudness change if the sound intensity ratio was 1000:1?
  4. What is the meaning of the reference intensity I0I_0 in the formula for loudness?
  5. How does the loudness relate to the physical energy of the sound waves?

Tip: In decibel calculations, the intensity ratio is often used to quantify loudness because the human ear perceives sound logarithmically, meaning a small increase in intensity can result in a large perceived increase in loudness.

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Math Problem Analysis

Mathematical Concepts

Logarithms
Sound Intensity
Decibels
Logarithmic Functions

Formulas

L(I) = 10 \log \left( \frac{I}{I_0} \right)
Delta L = 10 \log \left( \frac{I_{ad}}{I_{prog}} \right)

Theorems

Logarithmic Properties (log a - log b = log (a/b))

Suitable Grade Level

Grades 9-12