Intensity of sound – problems and solutions

Intensity of sound – problems and solutions

1. Point A and B located at 4 meters and 9 meters from a source of the sound. If I_A and I_B are intensity at point A and point B, then I_A : I_B =…

Known :

The distance of point A from a source of sound (r_A) = 4 meters

A distance of point B from the source of sound (r_B) = 9 meters

The intensity of sound at point A = I_A

The intensity of sound at point B = I_B

Wanted: I_A : I_B

Solution :

I_A r_A² = I_B r_B²

I_A 4² = I_B 9²

I_A 16 = I_B 81

I_A / I_B = 81/16

2. The intensity of a source of sound is 10⁻⁹ Wm⁻². I_o = 10⁻¹² Wm⁻². What is the sound level of 10 sources of sounds?

Solution :

Known :

I = 10^-9 W/m²

I_o = 10^-12 W/m²

x = 10

Known: Sound level (β)

Solution :

3. The sound level of a source of sound is 10 dB. What is the intensity of 1000 sources of sound? The minimum intensity I_o = 10⁻¹² Wm⁻².

Known :

β = 10 dB

I_o = 10^-12 W/m²

x = 1000

Wanted : Intensity

Solution :

The intensity of a source of sound :

The intensity of 1000 sources of sound :

I = (1000)(10^-11) = (10³)(10^-11)

I = 10^-8 W/m²

4. The sound level of A is 40 dB, and the sound level of B is 60 dB. I_o = 10^-12 W m^-2. Determine 100β_A : 10β_B.

Known :

The sound level of A = 40 dB

The sound level of B = 60 dB

I_o = 10^-12 W m^-2.

Wanted : 100β_A : 10β_B

Solution :

The sound level of 100 A :

β = 40 + 10 log 100

β = 40 + 10 log 10²

β = 40 + (2)(10)(log 10)

β = 40 + (2)(10)(1)

β = 40 + 20

β = 60 dB

The sound level of 10 B :

β = 60 + 10 log 10

β = 60 + 10 log 10¹

β = 60 + (1)(10)(log 10)

β = 60 + (1)(10)(1)

β = 60 + 10

β = 70 dB

β_A : β_B

60 : 70

6 : 7

5. If point P is the source of sound, then the ratio of sound intensity at points S, R, and Q is …

Known :

The distance of point Q from the sound source (r_Q) = 3 meters

The distance of point R from the sound source (r_R) = 6 meters

The distance of point S from the sound source (r_S) = 5 meters

Sound intensity at point Q = I_Q

Sound intensity at point R = I_R

Sound intensity at point S = I_S

Wanted: Comparison of sound intensity at points S, R, and Q (I_S: I_R: I_Q)

Solution :

Sound intensity at point S :

Sound intensity at point R :

Sound intensity at point Q :

Comparison of sound intensity at points S, R, and Q :

6. Point A is P from the sound source, point B is 2P from source sound and point C is 4P from the sound source. Comparison of sound intensity at A, B, and C is …

Known :

Distance point A from the sound source (r_A) = P

Distance point B from sound source (r_B) = 2P

Distance point C from sound source (r_C) = 4P

Sound intensity at point A = I_A

Sound intensity at point B = I_B

Sound intensity at point C = I_C

Wanted: Comparison of sound intensity at A, B, and C (I_A : I_B : I_C)

Solution :

Sound intensity at point A :

Sound intensity at point B :

Sound intensity at point C :

Comparison of sound intensity at A, B, and C :

7. A total of 100 people were singing. If the level of the sound intensity of one student when singing 40 dB (assuming for each child is the same), then the intensity of the resulting sound is … (I_o = 10^-12 W.m^-2)

Known :

Level of the intensity of one student (TI) = 40 dB

I_o = 10^-12 W/m²

The number of students (x) = 100

Wanted: The sound intensity of 100 students

Solution :

Sound intensity of a student :

Sound intensity of 100 students :

I_x = (x)(I)

I_x = (100)(10^-8) = (10²)(10^-8)

I_x = 10^-6 W/m²

8. The sound intensity of 100 identical machines is 10^-7 Watt.m^-2. If I_o = 10^-12 Watt.m^-2, then the level of the sound intensity of a machine is…

Known :

Sound intensity of 100 machines (I_x) = 10^-7 Watt.m^-2

I_o = 10^-12 W/m²

The number of machines (x) = 100 = 10²

Wanted : The sound intensity level of a machine (TI)

Solution :

Sound intensity of a machine :

I_x = (x)(I)

10^-7 = (10²)(I)

I = 10^-7 / 10² = (10^-7)(10^-2) = 10^-9

The level of sound intensity of a machine (TI) :

9. The sound intensity of a source of sound is 6 x 10^-6 W/cm². If the sound intensity level increased by 10 db, determine the intensity of sound.

Known :

Intensity (I) = 6 x 10^-6 W/cm²

I_o = 10^-12 W/m² = 10^-12 W / 10⁴ cm² = 10^-16 W/cm²

Wanted : The intensity of sound

Solution :

The addition of the sound intensity of 10 W/m² = 10^-3 W/cm² is equivalent to the addition of the sound intensity level of 10 dB. The addition of sound intensity of 10² W/m² = 10^-2 W/cm² is equivalent to the addition of sound intensity level of 20 dB. And so on.

If the intensity level is increased by 10 dB, the intensity increases by 10^-3 W/cm². So the intensity becomes 6 x 10^-6 W/cm²) + (10^-4 W/cm²) = (6 x 10^-6 W/cm²) + (100 x 10^-6 W/cm²) = 106 x 10^-6 W/cm².

10. Observer A is 5 m away from a sound source. While observer B is 10 m from the same sound source. So the ratio of sound intensity heard by observer B and A is…

Known :

The distance of A from the source of sound (r_A) = 5 meters

The distance of B from the source of sound (r_B) = 10 meters

Wanted: The ratio of sound intensity heard by observer B and A

Solution :

I_B : I_A = 1 : 4

The correct answer is B.

1. What is sound intensity and how is it different from loudness?
   • Answer: Sound intensity is a measure of the sound power per unit area, usually measured in watts per square meter (W/m²). Loudness, on the other hand, is a subjective perception of the strength of a sound, typically measured in decibels (dB). While intensity is an objective, physical measurement, loudness is a subjective experience that can vary between individuals.
2. How does the inverse square law relate to sound intensity?
   • Answer: The inverse square law states that the intensity of sound decreases with the square of the distance from the source. This means that as you move farther from the source, the intensity drops off rapidly.
3. Why do two sounds with the same intensity not necessarily have the same perceived loudness?
   • Answer: The human ear has different sensitivities to different frequencies. Therefore, even if two sounds have the same intensity but different frequencies, they might be perceived as having different loudness levels.
4. How does sound intensity relate to the amplitude of a sound wave?
   • Answer: Sound intensity is directly proportional to the square of the amplitude of the sound wave. If the amplitude of a sound wave is doubled, the intensity will increase by a factor of four.
5. Why might a whisper in a quiet room be clearly audible, but the same whisper next to a roaring waterfall be inaudible?
   • Answer: This is due to the difference in background noise levels. In a quiet room, the background noise level is low, allowing the whisper to stand out. Next to a waterfall, the intense noise masks the quiet whisper, making it inaudible.
6. What is the threshold of hearing and the threshold of pain in terms of sound intensity?
   • Answer: The threshold of hearing is the quietest sound that the average human ear can hear, typically at an intensity of about 10⁻¹² W/m². The threshold of pain, which represents sound levels that can cause pain and potential damage to the ear, is around 1 W/m².
7. Why do sounds seem quieter when heard through a wall or other barrier?
   • Answer: When sound waves encounter a barrier, some of the energy is reflected, absorbed, and transmitted. The absorption and reflection reduce the intensity of the sound that is transmitted through the barrier, making it sound quieter.
8. How does frequency affect the intensity of sound that is absorbed by materials?
   • Answer: Different materials have varying absorption coefficients for different frequencies. For instance, low frequencies (bass sounds) are harder to absorb and can penetrate through walls more easily than higher frequencies.
9. Why do concert venues and theaters invest in soundproofing and acoustic treatments?
   • Answer: Soundproofing helps to keep unwanted external noise out and prevents sound from leaking outside, ensuring that only the desired sound is heard within the venue. Acoustic treatments help in controlling sound reflections within the venue to ensure clear and uniform sound distribution.
10. How does the intensity of sound relate to potential hearing damage?

• Answer: Prolonged exposure to high-intensity sound levels can lead to hearing damage or loss. The intensity and duration of exposure both play roles; even short exposures to extremely loud sounds can cause permanent damage, while long-term exposure to moderately loud sounds can also be harmful.