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Cerrena [4.2K]
3 years ago
13

The Doppler effect is an effect produced by a moving source of sound or electromagnetic waves due to the relative motion of a so

urce or a receiver. If an emergency vehicle approaches Bob and moves away from Jill, how does the frequency of the siren change?
Physics
2 answers:
IrinaK [193]3 years ago
8 0

Answer:

Apparent frequency increases for Bob and decreases for Jill.

Explanation:

Doppler effect refers to the apparent change in frequency when there is relative motion between source and the listener. When the source and listener move closer, the apparent frequency heard by the listener increases. And when the source and listener move away from each other, the apparent frequency heard by the listener decreases.

Thus, when the emergency vehicle approaches Bob, the apparent frequency of the siren increases. It moves away from Jill, the apparent frequency  heard by the Jill decreases.

mojhsa [17]3 years ago
5 0

Answer:

change in frequency

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A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr
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Answer:

<u>Part A</u>

I = 1.4 mW/m²  

<u>Part B</u>

β = 91.46 dB

Explanation:

<u>Part A</u>

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

                                            I = \frac{P}{A}

                                            I = \frac{P}{4\pi r^{2}}

Where;

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I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = \frac{34}{4\pi (44)^{2}}

                                     I = 0.0013975 W/m²

                                 ≈  I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

                                     I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

<u>Part B</u>

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value.  It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;  

                                    β = 10log_{10}\frac{I}{I_{0}}  dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

           ∴        β = 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}

                      β = 10log_{10} (1.4 * 10^{9})

                      β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.                

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