Question

A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as 10⁻¹²W/m²]
(A) 30 cm (B) 10 cm
(C) 40 cm (D) 20 cm

Answer 1

Answer:

c answer

is the correct answer ok

Answer 2

A small speaker delivers 2 W of audio output. At a distance of (C) 40 cm from the speaker will one detect 120 dB intensity sound.

• Given:

Radiated power : P = 2 W

Sound intensity in dB = I$_{dB}$ = 120 dB

Reference intensity of sound: Io = 10⁻¹² W/m²

• Sound intensity (Loudness) in dB is given as

$I_{dB} = \log_{10}(\frac{I}{Io} )$ ......................(1)

where I is the intensity of sound in W/m²

• Solving equation (1) for I, we get

$120 = 10\ log_{10}(\frac{I}{10^{-12}} )\\\\12 = log_{10}(\frac{I}{10^{-12}} )\\\\I = 10^{12}\times 10^{-12}\\\\I = 1\ W/m^2$......................(2)

• Intensity of sound is given as the ratio of power radiated per unit area.

• $I = \frac{P}{A}$ .............................(3)

• The power radiated from speaker (sound energy from the source) is in a spherical form and hence A is equal to the area of sphere (A = 4$\pi$R²) with radius R.

• This denotes the intensity of sound at a distance R from the speaker.

From (2) and (3), we get

$I = \frac{P}{A}\\\\I = \frac{P}{4\pi R^2}\\\\1 = \frac{2}{4\times \pi \times R^2}\\\\R^2 = \frac{1}{2\pi } = 0.1591 \m^2\\\\R = \sqrt{0.1591} = 0.398\ m\\\\0.398\m = 39.8\ cm$

• 39.8 cm ≈ 40 cm