Question

A child sets off the firecracker at a distance of 100 m from the family house. what is the sound intensity β100 at the house?

Answer 1

To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

 

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

 

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

 

Answer:

100 dB

Answer 2

The intensity of the sound produced due to the firecracker at the distance of $100\text{ m}$ is $\fbox{\begin\\100\text{ dB}\end{minispace}}$.

Further Explanation:

The vibration, produced in the medium, which travels as an wave of pressure or density through a medium is known as sound. The sound is a longitudinal wave and it requires a medium for its propagation.  

Given:

The distance between the place where firecracker is set off by a child and the family house is $100\text{ m}$.  

The intensity of the sound produced by the firecracker at the distance $1\text{ m}$ is$140\text{ dB}$.

Concept:

The intensity of the sound wave is defined as the power carried by the sound waves in the direction perpendicular to the direction of propagation per unit time.

The intensity of the sound wave is:

$\fbox{\begin\\I=\dfrac{P}{4\pi r^2}\end{minispace}}$                                                        ...... (1)

Here, $I$ is the intesnity of the sound wave, $P$ is the power carried by the sound wave and $r$ is the distance between source and the listener.

The intensity of the sound in decible at the house is:

$\fbox{\begin\\\beta _{100}=\beta _{1}+10\log \dfrac{I_{100}}{I_{1}}\end{minispace}}$ ...... (2)

Here, $\beta _{100}$ is the intensity of the sound at the distance of $100\text{ m}$, $\beta _{1}$ is the intensity of sound produced by the firecracker at the distance of $1\text{ m}$, $I_{100}$ is the intensity of sound produced by firecracker at $100\text{ m}$ and $I_{1}$ is the intensity of sound produced by firecracker at $1\text{ m}$.

The ratio of the intensity of sound at distance $100\text{ m}$ and $1\text{ m}$ is:

$\fbox{\begin\\\dfrac{{{I_{100}}}}{{{I_1}}}=\frac{{r_1^2}}{{r_2^2}}\end{minispace}}$                                                                                     …… (3)

Calculation:

Substitute the values in equation (3).

$\begin{aligned}\frac{{{I_{100}}}}{{{I_1}}}&=\frac{{{{\left( 1 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\\&={10^{ - 4}}\\\end{aligned}$

Substitute the values in equation (2).

$\begin{aligned}{\beta _{100}}&=140\,{\text{dB}} - 40\,{\text{dB}} \\&=100\,{\text{dB}}\\\end{aligned}$.

Thus, the intensity of the sound produced due to the firecracker at the distance of $100\text{ m}$ is $\fbox{\begin\\100\text{ dB}\end{minispace}}$

Learn more:

1.  The motion of a body under friction brainly.com/question/4033012

2.  A ball falling under the acceleration due to gravity brainly.com/question/10934170

3. Conservation of energy brainly.com/question/3943029

Answer Details:

Grade: College

Subject: Physics

Chapter: Waves and Oscillation

Keywords:

Intensity of sound, sound waves, vibration, firecracker, power carried by sound waves, 100db, 100 dB, 100dB, 100 m, bursting of firecrackers, sound inside the house, 140 db, 140 dB.