Basic Physics

# 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 IA and IB are intensity at point A and point B, then IA : IB =…

Known :

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

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

The intensity of sound at point A = IA

The intensity of sound at point B = IB

Wanted: IA : IB

Solution :

IA rA2 = IB rB2

IA 42 = IB 92

IA 16 = IB 81

IA / IB = 81/16

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

Solution :

Known :

I = 10-9 W/m2

Io = 10-12 W/m2

x = 10

Known: Sound level (β)

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

Known :

β = 10 dB

Io = 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. Io = 10-12 W m-2. Determine 100βA : 10βB.

Known :

The sound level of A = 40 dB

The sound level of B = 60 dB

Io = 10-12 W m-2.

Wanted : 100βA : 10βB

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

βA : βB

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 (rQ) = 3 meters The distance of point R from the sound source (rR) = 6 meters

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

Sound intensity at point Q = IQ

Sound intensity at point R = IR

Sound intensity at point S = IS

Wanted: Comparison of sound intensity at points S, R, and Q (IS: IR: IQ)

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 (rA) = P

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

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

Sound intensity at point A = IA

Sound intensity at point B = IB

Sound intensity at point C = IC

Wanted: Comparison of sound intensity at A, B, and C (IA : IB : IC)

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 : Read :  Centripetal force in uniform circular motion – problems and solutions

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 … (Io = 10-12 W.m-2)

Known :

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

Io = 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 :

Ix = (x)(I)

Ix = (100)(10-8) = (102)(10-8)

Ix = 10-6 W/m2

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

Known :

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

Io = 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 :

Ix = (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

Io = 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 (rA) = 5 meters

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

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

Solution : IB : IA = 1 : 4