Define sound intensity - Leaving Cert Physics - Question b - 2007

To find the sound intensity at a distance of 3 m from a loudspeaker with a power rating of 25 mW, we first calculate the surface area over which the sound is dispersed. Assuming spherical spreading:

$A = 4\pi r^2$

For $r = 3 m$:

$A = 4\pi (3)^2 = 36\pi \approx 113.1 \, m^2$

Now, using the power rating:

$I = \frac{P}{A} = \frac{25 \times 10^{-3}}{36\pi} \approx 2.21 \times 10^{-4} \, W/m^2$

In the sound intensity? The new loudspeaker has a power of 50 mW. The sound intensity can be calculated in the same way:

$I_{new} = \frac{P_{new}}{A} = \frac{50 \times 10^{-3}}{36\pi} \approx 4.42 \times 10^{-4} \, W/m^2$

Therefore, the change in sound intensity is:

$\Delta I = I_{new} - I = 4.42 \times 10^{-4} - 2.21 \times 10^{-4} = 2.21 \times 10^{-4} \, W/m^2$

In the sound intensity level? The sound intensity levels can be calculated using the formula:

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

where $I_0 = 1 \times 10^{-12} \, W/m^2$ is the reference intensity.

For the original intensity:

$L_{original} = 10 \log_{10} \left( \frac{2.21 \times 10^{-4}}{1 \times 10^{-12}} \right) \approx 10 \log_{10} (2.21 \times 10^{8}) \approx 83.44 \, dB$

For the new intensity:

$L_{new} = 10 \log_{10} \left( \frac{4.42 \times 10^{-4}}{1 \times 10^{-12}} \right) \approx 10 \log_{10} (4.42 \times 10^{8}) \approx 86.46 \, dB$

Thus, the change in sound intensity level is:

$\Delta L = L_{new} - L_{old} = 86.46 - 83.44 = 3.02 \, dB$.

The human ear's sensitivity varies across different frequencies. This is often accounted for by using A-weighted decibels (dBA), which adjusts the sound intensity levels to reflect the frequency response of human hearing. This scale emphasizes frequencies between 2 kHz and 4 kHz, which are critical for human speech understanding.