You are standing 1 meter from a squawking parrot. If you move to a distance of 3 meters away, the sound intensity will be what fraction of its original value?

Use the inverse square law, thus if you move a distance of 3m away, the sound intensity decrease by 1/3^2= 1/9

Answered by Anonymous | 2024-06-10

If you move to a **distance **of three **meters **away, the sound intensity will be **1/9 **of its original value.
What is the intensity of sound?
The subjective experience of sound pressure is called its intensity . Since everyone **experiences **loudness differently, loudness cannot be measured scientifically. Thus, the relative nature of the **acoustic **sound intensity, audio intensity, or acoustic signal.
**Decibels **are the units used to measure sound intensity, often known as sound power or sound pressure . Alexander Graham Bell, the creator of both the telephone and the audiometer, is remembered by the name of decibels (dB).
The **intensity **of **sound **is inversely proportional to the **distance **between the source and the observer.
Intensity **proportional **to k/ro²
I = k/ro²
Where ro² = 1 meter
I₁ = K
r = 3 meter
I₃ = K/32 = K/9 = I₁/9
The sound intensity will be 1/9 of the original value.
Therefore, the **sound intensity **will be 1/9 of its original value.
To learn more about intensity , refer to the link:
https://brainly.com/question/15089336
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Answered by mahima6824soni | 2024-06-16

As you move from 1 meter to 3 meters away from a sound source, the sound intensity decreases to 9 1 ​ of its original value. This decrease in sound intensity is due to the inverse square law. The formula used to determine this is I ∝ r 2 1 ​ .
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Answered by mahima6824soni | 2024-10-02