# Sound Intensity Formula

Sound intensity is defined as the amount of sound per unit area perpendicular to the route of the sound waves, which is denoted by ‘**I’**. The standard definition is the measurement of sound energy as a quantity of noise sound intensity in the air at the listener position. The SI unit of sound intensity is watt per square meter (W/m2).

### Formula for Sound Intensity

The sound intensity formula can be defined as the ratio of the power of the sound wave to the area.

I = P/Awhere,

- P = power of the sound
- A = Area

### Sample Problems

**Problem 1: A person is cheering at a music concert at a power of 2×10 ^{-4} W. Calculate the sound intensity at a distance of 10m.**

**Solution:**

Given,

- P (power) = 2×10
^{-4}- A (here distance) = 10 m
So by using the sound intensity formula,

I = P / AI = 2×10

^{-4}/ 10

I = 2×10^{-5}

Thus the sound intensity in this condition will be 2×10^{-5}W/m^{2}

**Problem 2: A student at the last bench is 20m away from his teacher and he asks doubt at a sound intensity of 1.1×10 ^{-4}. Find at what power he has asked the doubt so that his teacher heard his doubt?**

**Solution:**

Given,

- sound intensity (I): 1.1×10
^{-4}- Distance (A) : 20m
So by rearranging the formula for sound intensity,

P = I × A

and substituting the values of I and A,

P = 1.1 ×10

^{-4}×20

P = 2.2×10^{-3}

Thus he has asked the doubt at a power of 2.2×10^{-3}W

**Problem 3: What will be the intensity of the siren of an Ambulance with a power of 8×10 ^{-4} at a distance of 30m?**

**Solution:**

Given,

- power (P) = 8×10
^{-4}- Distance (A) = 30m
So, according to the sound intensity formula,

I = P / AI = 8×10

^{-4}×30

I = 0.024

Thus the sound intensity in this condition will be 0.024 W/m^{2}

**Problem 4: What will be the distance between two friends in a hall if the one friend calls him at a sound power of 5×10 ^{-4} at 10×10^{-4} sound intensity?**

**Solution:**

Given,

- Power (P) = 5×10
^{-4}- Sound intensity (I) = 10×10
^{-4}So by rearranging the formula for sound intensity,

A = P / Iand substituting the values of I and P,

A = 5×10

^{-4}/ 10×10^{-4}

A = 0.5m

Thus the distance between them is 0.5m.

**Problem 5: What is the** **sound intensity of a singer who is singing at a** **power of 6×10 ^{-4} relative to a person at a distance of 3m.**

**Solution:**

Given,

- Power (P) = 6×10
^{-4}- Distance (A) = 3×10
^{-4}So, according to the sound intensity formula,

I = P / AI = 6×10

^{-4}/ 3

I = 2×10^{-4}

Thus the sound intensity in this condition will be 2×10^{-4}W/m^{2}