8. (a) Destructive interference can occur when waves from coherent sources meet - Leaving Cert Physics - Question 8 - 2011

1. The waves must be coherent, meaning they maintain a constant phase difference over time.
2. The waves must have the same frequency to ensure they oscillate synchronously.

P: node Q: antinode

To find the frequency, we use the wave speed formula: v = f * λ

Here, we know the length of the pipe is 90 cm, which corresponds to a quarter wavelength for the fundamental mode. Therefore:

λ = 4 * length = 4 * 0.90 m = 3.60 m

Now substituting into the wave speed formula:

f = rac{v}{λ} = rac{340 ext{ m/s}}{3.60 ext{ m}} ext{ } ightarrow f ext{ } ext{ } ext{ } ext{ } ext{ } = 94.44 ext{ Hz}

For a pipe closed at one end, the fundamental frequency is given by: f_{fundamental} = rac{v}{4L} ext{ } where L is the length of the pipe:

In this case: L = 0.90 m

Therefore, f_{fundamental} = rac{340 ext{ m/s}}{4 * 0.90 ext{ m}} = 94.44 ext{ Hz}

A clarinet produces odd harmonics since it is a pipe closed at one end. The harmonics produced are in the form of 1st (fundamental), 3rd, 5th, and so on.

Using the formula for sound intensity:

$I = \frac{P}{A}$

where A is the area of a sphere given by $A = 4\pi R^2$,

For a spherical wave at a distance of 8 m:

$A = 4\pi (8^2) = 4\pi (64) = 256\pi$

Thus, $I = \frac{100}{256\pi} \approx 0.124 \text{ W/m²}$.

The relationship between sound intensity levels can be expressed as:

$L_{S1} - L_{S2} = 10 log_{10} \left( \frac{I_1}{I_2} \right)$,

Where a change of 3 dB indicates halving the intensity, thus: $I_2 = \frac{I_1}{2}$

To find the new distance R where intensity is reduced by 3 dB:

$I_2 = \frac{100}{4\pi R^2} = \frac{100}{2} = \frac{50}{4\pi R^2}$

After some calculations, the new distance R is found to be approximately: R = 11.13 m.