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3 years ago

when a sound's intensity increases by a factor of 10, its decibel reading increases by how many decibels?

Physics

2 answers:

Anit [1.1K]3 years ago

8 0

Answer:

If sound intensity increases by a factor of 10, the decibel reading increases by 10.

Explanation:

Decibel scale:

In order to express levels of sound meaningfully in numbers that are more manageable, a logarithmic scale is used, rather than a linear one. This scale is the decibel scale. every increase of 3 dB represents a doubling of sound intensity.

Increase in decibel scale is given by the formula:

n = 10*log(Increased Intensity of sound/Original Intensity of Sound

n = 10*log(I/I₀)

In our case the sound intensity increases by a factor of 10. So I/I₀ increased by 10 times and will be 10 I/I₀

n = 10*log(10 I/I₀)

n = 10*log(10) + 10log(I/I₀)

n = 10 + 10*log(I/I₀)

This shows that the decibel reading is increased by 10 points.

PIT_PIT [208]3 years ago

4 0

Answer:

Explanation:

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3 years ago

A violin string has a length of 327mm and produces a note of frequency 440Hz.

Scorpion4ik [409]

The characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is:

The new frequency is f = 657 Hz

<h3>How is a standing wave produced?</h3>

A standing wave is produced when a traveling wave meets an obstacle and bounces, the sum of the two waves results in a wave that does not propagate in space.

In the event that the obstacle is a fixed point, there is a node at this point. The expression for the length of the standing wave.

L = $\frac{\lambda }{2}$ fundamental frequency

L = $2 \frac{\lambda}{2}$ second harmonic

L = $3 \frac{\lambda}{2}$ third harmonic

L = $n \frac{\lambda}{2}$ general term.

Where L is the length of the chord, lan the wavelength and n an integer.

Wave speed is related to wavelength and frequency.

v = λ f.

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v = $\frac{2L}{n}$

They indicate that initially the string has a length of L₀ = 327 mm= 0.327m and the frequency is f₀ = 440 Hz.

v n = 2L₀ f₀

v n = 2 0.327 440

v n = 287.76

They indicate that the tension on the string do not changes and the speed of the wave depends only on the tension and the density of the string, therefore it is constant, we assume that the harmonic does not change either, therefore the new length.

v n = 2 L f

Let's substitute.

287.76 = 2 L f

f = $\frac{287.76x}{2L}$

Let's calculate.

f = $\frac{287.76}{2 \ 0.219}$

f = 656.99 Hz

In conclusion with the characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is:

The new frequency is: f = 657 Hz

Learn more about standing waves here: brainly.com/question/17031219

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when a sound's intensity increases by a factor of 10, its decibel reading increases by how many decibels?​

when a sound's intensity increases by a factor of 10, its decibel reading increases by how many decibels?