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Lena [83]
3 years ago
7

The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

oes the intensity level drop to the safe value of 85 dB?
Physics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

28.11m far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

  (1). \: \:\beta =10 log(\dfrac{I}{I_0})

The ratio of the intensities can be written as

$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $

\dfrac{I}{I_0} = \dfrac{A_0}{A}.

And since

A = 4\pi r^2

and

A_0 = 4 \pi r_0^2,

\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2}  = \dfrac{r_0^2}{r^2}

meaning

\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.

Putting this into equation (1), we get:

\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

100dB=10 log(\dfrac{r_02}{(5m)^2})

10= log(\dfrac{r_0^2}{25})

by taking both sides to the exponent:  

10^{10}= \dfrac{r_0^2}{25}

r^2 = 25 *10^{10}\\r = 5 *10^5

Now equation (2) becomes

\beta = 10log(\dfrac{25*10^{10}}{r^2})

when the intensity level is 85 dB we have

85 = 10log(\dfrac{25*10^{10}}{r^2})

8.5 = log(\dfrac{25*10^{10}}{r^2})

take both sides to exponents and we get:

10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}

10^{8.5} =\dfrac{25*10^{10}}{r^2}

r^2 = \dfrac{25*10^{10}}{10^{8.5}}

\boxed{r = 28.11m}

Thus, 28.11m far from the speaker the intensity drops to 85 dB.

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