# What is the total acoustical power output of the siren?

• Jayhawk1
In summary, the problem is that I can not find the dB level at a distance of 21.5 m from the siren. The total acoustical power output of the siren is 1100 watts. The sound intensity is reduced by 15 dB from its level at 15 m.
Jayhawk1
I tried getting some help before... but I couldn't figure out the logs- if anyone could help me (especially with the manipulation of the logs) I'd appreciate it. Here's the problem:

11) [3.0/4.0] A tornado warning siren on top of a tall pole radiates sound waves uniformly in all directions. At a distance of 15 m the sound intensity of the siren is 0.39 . Neglect any effects from reflection of the sound waves from the ground. a) At what distance from the siren is the intensity 0.19 ? b) What is the total acoustical power output of the siren? c) At what distance is the sound intensity reduced by 15 dB from its level at 15 m?

Now I did get parts A and B:

a) 21.5 m
b) 1100 Watts

The decibels of a magnitude is defined in relation to some base 0 dB level:

$$V(dB) = 20 log(V/V_0)$$

But this will cancel out when finding the difference between the the decibel level of two magnitudes:

$$V_2(dB) - V_1(dB) = 20 log(V_2/V_0) - 20 log(V_1/V_0) = 20 log((V_2/V_0)\cdot(V_0/V_1))=20 log(V_2/V_1)$$

You have the equation for this, I know. When you have something in a log that you have to figure out, you can eliminate the log by isolating it and using both side of the equation as a power of 10. The log is the inverse of 10 to a power

a = b log(c)
(a/b) = log(c)
10^(a/b) = c

Use this on the db level equation you already have

Specifically,

$$L = 10log({\frac{I}{I_0}})$$
$$-15/10 = log({\frac{I}{I_0}})$$
$$10^{-1.5} = {\frac{I}{I_0}}$$
$$I_0 10^{-1.5} = I$$

You have the intensity at one point $$I_0$$. Now you can find it at the second point. Then you can calculate the distance you need to find.

The factor 20 or 10 in the dB equation depends on whether you are talking about amplitude or energy. In your case, you are working with intensity (energy/area) so the factor is 10.

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This doesn't make any sense to me. How does intensity work into the equation?

From above

$$I_0 10^{-1.5} = I$$

$$.39 * 10^{-1.5} = I$$

$$.0123 = I$$

Now it is just like part a)

Last edited:

## 1. What is the definition of acoustical power output?

The acoustical power output of a device measures the amount of sound energy that is emitted by the device in a given amount of time. It is typically measured in watts (W) or decibels (dB).

## 2. How is the total acoustical power output of a siren calculated?

The total acoustical power output of a siren is calculated by measuring the sound intensity at multiple points around the siren and then summing up the intensities to get the total power output. This can be done using specialized equipment such as a sound level meter.

## 3. What factors affect the total acoustical power output of a siren?

The total acoustical power output of a siren can be affected by several factors such as the size and design of the siren, the power source, and the environment in which it is used. The materials used in the construction of the siren can also impact its power output.

## 4. How does the frequency of a siren affect its acoustical power output?

The frequency of a siren can have a significant impact on its acoustical power output. Generally, higher frequencies require more power to produce the same sound intensity as lower frequencies. Therefore, a siren with a higher frequency may have a higher acoustical power output compared to one with a lower frequency.

## 5. Can the acoustical power output of a siren be increased?

Yes, the acoustical power output of a siren can be increased by using a more powerful power source or by optimizing the design and materials used in the construction of the siren. However, it is important to consider the potential impact on noise pollution and adhere to any regulations regarding sound levels in the area where the siren will be used.

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