The Fall of the Portable Radio: How Long for a 10 dB Increase?

[Salty]

A portable radio is sitting at the edge of a balcony 5.1m above the ground. The unit is emitting a sound uniformly in all directions. By accident, it falls from rest off the balcony and continues to play all the way down. A gardener is working in a flower bed directly below the falling unit. From the instant the unit begins to fall, how much time is required for the sound intensity level heard by the gardener to increase by 10 dB?

 

[MathStudent]

hint: first find the distance from the radio the gardener would have to be to here an increase in intensity of 10dB

A useful equation is the inverse square law for intensity

$$\frac{I_1}{I_2} = (\frac{r_2}{r_1})^2$$

where I1 is the intensity at a distance r1, and I2 is the intensity at a distance r2.

 

[thepatientmental]

The fall of the portable radio is a common occurrence that can happen to anyone. In this scenario, the radio is emitting sound uniformly in all directions and is sitting at the edge of a balcony 5.1m above the ground. It is important to note that sound intensity decreases with distance according to the inverse square law, which states that the intensity is inversely proportional to the square of the distance from the source.

As the radio falls, it will continue to emit sound at the same intensity, but the distance between the radio and the gardener will decrease. This means that the sound intensity level heard by the gardener will increase as the radio gets closer to the ground. The gardener will not hear any change in sound intensity until the radio reaches a distance of 5.1m from the ground. At this point, the sound intensity level heard by the gardener will increase by 10 dB.

To calculate the time required for the sound intensity level to increase by 10 dB, we can use the formula for sound intensity level (L) in decibels:

L = 10 log (I/I0)

Where I is the sound intensity and I0 is the reference intensity (10^-12 W/m^2).

Since we know that the sound intensity level will increase by 10 dB, we can set L equal to 10 dB and solve for the time (t):

10 = 10 log (I/I0)

1 = log (I/I0)

10 = I/I0

I = 10^-2 I0

Using the inverse square law, we can set the initial distance (d1) to 5.1m and the final distance (d2) to 0m (when the radio reaches the ground):

I2/I1 = (d1/d2)^2

I2 = I1(d1/d2)^2

I2 = (10^-2 I0)(5.1/0)^2

I2 = (10^-2 I0)(26.01)

I2 = 0.2601 I0

Now we can set I2 equal to the reference intensity (10^-12 W/m^2) and solve for the time (t):

0.2601 I0 = 10^-12 W/m^2

I0 = 10^-12 W/m^2/0.2601

I0 = 3.844 x 10^-11 W/m