We can solve this problem using the concept of sound intensity level, which is measured in decibels (dB). The formula for calculating the sound intensity level $L$ in decibels is:

$L = 10 \log_{10} \left( \frac{I}{I_0} \right)$

where:

• $I$ is the sound intensity in $\text{W/m}^2$,
• $I_0$ is the reference intensity, which is $1 \times 10^{-12} \, \text{W/m}^2$ (the threshold of hearing).

(a) Sound intensity level for one speaker:

The intensity of the sound wave from one speaker is given as $I = 1.8 \times 10^{-7} \, \text{W/m}^2$.

Using the formula:

$L = 10 \log_{10} \left( \frac{1.8 \times 10^{-7}}{1 \times 10^{-12}} \right)$

Let's calculate this value.

(b) Sound intensity level for both speakers:

When both speakers are turned on, the total intensity doubles because the sound waves are of the same intensity. So the total intensity $I_{\text{total}}$ is:

$I_{\text{total}} = 2 \times 1.8 \times 10^{-7} = 3.6 \times 10^{-7} \, \text{W/m}^2$

Now, using the same formula for sound intensity level:

$L_{\text{total}} = 10 \log_{10} \left( \frac{3.6 \times 10^{-7}}{1 \times 10^{-12}} \right)$

I'll calculate both these values for you.### (a) The sound intensity level heard by the listener when only one speaker is turned on is approximately 52.55 dB.

(b) The sound intensity level heard by the listener when both speakers are turned on is approximately 55.56 dB.

This shows that doubling the intensity (by turning on both speakers) increases the sound intensity level by about 3 dB.

Would you like more details or have any questions?

Here are some related questions:

1. What is the relationship between sound intensity and sound intensity level in decibels?
2. Why does doubling the intensity result in a 3 dB increase?
3. How is the reference intensity $I_0$ determined in acoustic measurements?
4. How would the intensity level change if the speakers were not at the same distance from the listener?
5. What other factors affect the perception of sound besides intensity?

Tip: The logarithmic nature of the decibel scale makes it more practical to represent large ranges of sound intensity compactly.