Numerical Problems 11.1: A normal conversation involves sound intensities of about 3.0 𝘅 10⁻⁶ Wm⁻². What is the decibel level for this intensity? What is the intensity of the sound for 100 dB? Ans. (64.8 dB, 0.01Wm-2)

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Part 1: To find the decibel level for the intensity of 3.0 𝘅 10⁻⁶ Wm⁻².?


Given Data:


Intensity of the sound ( I ) = 3.0 𝐱 10⁻⁶ w m⁻²

Intensity of faintest audible  sound ( I₀ ) = 10⁻¹² wm⁻²


Required:


Intensity Level = L - L₀ = ?



Solution:


We have the formula for intensity level as following



L  − L₀ = 10 log `\frac {I}{I₀}` dB
 
by putting corresponding values 

L  − L₀ = 10 log `\frac {3.0 ｘ 10⁻⁶ w m⁻²}{10⁻¹² wm⁻²}` dB

w m⁻² will cancel with each other So,

L  − L₀ = 10 log ( 3 ｘ 10⁻⁶⁺¹²) dB

L  − L₀ = 10 log ( 3 ｘ 10⁶) dB

L  − L₀ = 10 ( log (3) + log(10⁶) dB

L  − L₀ = 10 ( log (3) + 6 log10) dB

L  − L₀ = 10 ( 0.47 + 6ｘ 1) dB

L  − L₀ = 10 ( 0.47 + 6) dB

L  − L₀ = 10 (6.47) dB

L  − L₀ = 64.7 dB---------------Ans.1


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Part 2: To find the the intensity of the sound for 100 dB intensity level


Given:


Intensity Level = L - L₀ = 100 dB

Intensity of faintest audible sound ( I₀ ) = 10⁻¹² w m⁻²

To Find:


Intensity of the sound ( I ) = ? w m⁻²



Solution: 


We have the formula for intensity level as following

L  − L₀ = 10 log `\frac {I}{I₀}` dB

by putting values of each given term

100 dB = 10 log `\frac {I}{10⁻¹² w m⁻²}` dB

by dividing both side by 10 dB we get

10 = log `\frac {I}{10⁻¹² w m⁻²}`

Now taking antilog both side 

Antilog 10 = Antilog log `\frac {I}{10⁻¹² w m⁻²}`


10¹⁰ = `\frac {I}{10⁻¹² w m⁻²}` 

or


l  = 10¹⁰ ｘ 10⁻¹² w m⁻²

l  = 10¹⁰⁻¹² w m⁻²

l  = 10⁻² w m⁻²

or

l  = 0.01 w m⁻² --------------Ans.2  



Numerical Problems 11.2: If at Anarkali Bazar Lahore, intensity level is 80 dB, what will be the intensity of sound there? Ans. (10⁻⁴ Wm⁻²)

Ans. (1.63 msᐨ²)

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Given Data:


Intensity Level = L - L₀ = 80 dB

Intensity of faintest audible  sound ( I₀ ) = 10⁻¹² w m⁻²



Required:


Intensity of the sound ( I ) = ? wm⁻²



Solution: 


We have the formula for intensity level as following

L  − L₀ = 10 log `\frac {I}{I₀}` dB

by putting values of each given term

80 dB = 10 log `\frac {I}{10⁻¹² w m⁻²}` dB

by dividing both side by 10 dB we get

8 = log `\frac {I}{10⁻¹² w m⁻²}`

Now taking antilog both side 

Antilog 8 = Antilog log `\frac {I}{10⁻¹² w m⁻²}`


10⁸ = `\frac {I}{10⁻¹² w m⁻²}` 

or


l  = 10⁸ ｘ 10⁻¹² w m⁻²

l  = 10⁸⁻¹² w m⁻²

l  = 10⁻⁴ w m⁻²

or

l  = 0.0001 w m⁻² --------------Ans.



Numerical Problems 11.3: At a particular temperature, the speed of sound in air is 330 ms-1. If the wavelength of a note is 5 cm, calculate the frequency of the sound wave. Is this frequency in the audible range of the human ear? Ans. (6.6 x 10³ Hz, Yes)

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Given Data:


The Velocity of Sound ( V ) = 330 m s⁻¹ 

Wavelength ( λ ) = 5 cm = 0.05 m (conversion to SI unit)



Required:


Frequency ( f ) = ? Hz

 

Solution:


We know that Velocity (V), Frequency (f) and Wavelength (λ) are related by 

v = f λ 

f = `\frac{v}{λ}`

f = `\frac{330 m s⁻¹}{0.05 m}`

f = 6,600 s⁻¹       ∴{as f = `\frac{1}{T}` so, s⁻¹ = Hz}

f = 6.600 Hz

or

f = 6.6 𝐗 10³ Hz ---------------Ans.

Yes, f = 6,600 Hz lies in the range of human audible frequency range that is ranging from 20 Hz to 20,000 Hz.


Numerical Problems 11.4: A doctor counts 72 heartbeats in 1 min. Calculate the frequency and period of the heartbeats. Ans. (1.2 Hz, 0.83 s)

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Given Data:


Number of Heartbeat counts ( n ) = 72

Time ( t ) = 1 min = 60 sec (conversion: 1 m = 60 s)


Required:


Frequency of heartbeat ( f ) = ? Hz 

Time period of heartbeat ( t ) = ? sec



Solution:


We know that Frequency (f) of heartbeat is the number of heartbeat in one second, so

f = `\frac{Heartbeat count}{time}`

f = `\frac{72}{60 s}`

f = 1.2 s⁻¹            ∴{as f = `\frac{1}{T}` so, s⁻¹ = Hz}

f = 1.2 Hz --------------Ans.1

The frequency of the heart beat is 1.2 Hz


Now to find Time period we know that

T = `\frac{1}{f}`

So,

T = `\frac{1}{1.2 Hz}`

T = 0.833 s   -------------Ans.2

The time period of the heart beat is 0.833 sec 




Numerical Problems 11.5: A marine survey ship sends a sound wave straight to the seabed. It receives an echo 1.5 s later. The speed of sound in sea water is 1500 ms⁻¹. Find the depth of the sea at this position. Ans. (1125 m)

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Given Data:


Time (t)  taken by the sound wave = 1.5 sec

The speed of Sound in sea water ( v ) = 1500 ms⁻¹ 


Required:


Depth of the sea ( h ) = ? meters


Solution:


We will first find the total distance (S) traveled by the sound wave by an equations 

S = vt 

S =1500 m s⁻¹ 𝐱  1.5 s

S = 2250 m

∴{as this is the distance traveled by sound wave from ship (sea surface) to the seabed (bottom)  and then as a echo from the seabed to the ship (sea surface). So, depth of sea is the distance travelled by echo only and can calculated as a half of this total distance (S). thus,

h = `\frac{S}{2}`

h = `\frac{2250 m}{2}` 

h = 1125 m  ----------------Ans

Thus, Depth of the sea is 1125 meters 

Numerical Problems 11.6: A student clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the student if the speed of the sound is taken as 346 ms⁻¹. Ans. (865 m) 

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Given Data:


Total Time (t)  taken by the sound wave = 5 s 

The speed of Sound ( v ) = 346 m s⁻¹ 



Required:


Distance between the student and a cliff  ( d ) = ? meters



Solution:


We will first find the total distance (S) traveled by the sound wave travelled from student to cliff and then from cliff to student as an echo, by an equations 

S = vt 

S = 346 m s⁻¹ 𝐱  5 s

S = 1730 m

∴{as this is the distance traveled by sound wave from a student to the cliff  and then as a echo from the cliff to student is twice. So, for hearing echo distance (d) between the student and cliff must be half of this total distance (S). thus,

d = `\frac{S}{2}`

d = `\frac{1730 m}{2}` 

d = 865 m ---------------Ans.

Thus, distance between the cliff and a student is 865 meters 



Numerical Problems 11.7: A ship sends out ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 ms-1, what is the distance of the seabed from the ship? Ans. (2618 m)

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Given Data:


Time (t)  taken by the sound wave = 3.42 s

The speed of Ultrasound in sea water ( v ) = 1531 m s⁻¹ 


Required:


Depth of the sea ( h ) = ? meters



Solution:


We will first find the total distance (S) traveled by the Ultrasound wave by an equations 

S = vt 

S =1531 m s⁻¹ 𝐱  3.42 s

S = 5236.02 m

∴{as this is the distance traveled by Ultrasound wave from ship (sea surface) to the seabed (bottom)  and then as a echo from the seabed to the ship (sea surface). So, depth of sea is the distance travelled by echo only and can calculated as a half of this total distance (S). thus,

h = `\frac{S}{2}`

h = `\frac{5236.02 m}{2}` 

h = 2618 meters

Thus, Depth of the sea is 2618 meters 



Numerical Problems 11.8: The highest frequency sound humans can hear is about 20,000 Hz. What is the wavelength of sound in air at this frequency at a temperature of 20 ⁰C? What is the wavelength of the lowest sounds we can hear of about 20 Hz? Assume the speed of sound in air at 20 ⁰C is 343 ms⁻¹. Ans. (1.7 𝐱 10⁻² m, 17.2 m)

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Given Data:


The highest frequency of sound = f₁ = 20,000 Hz

The lowest frequency of sound = f₂ = 20 Hz

Speed of sound = v = 343 m s⁻¹ 


Required:


Wavelength of sound of highest frequency  = λ₁ = ? m

Wavelength of sound of lowest frequency  = λ₂ = ? m


Solution:


General Relation among Velocity (V), Frequency (f), and Wavelength (λ) are 

v = f λ

λ = `\frac{v}{f}`

For highest frequency sound the above equation will become as

λ₁ = `\frac{v}{f₁}`     ∴{as v₁ =v₂ =v}

λ₁ = `\frac{343 m s⁻¹}{20000 Hz }` 

∴{as f = `\frac{1}{T}` so, s⁻¹ = Hz}

λ₁ = `\frac{343 m s⁻¹}{20000 s⁻¹ }`  

λ₁ = 0.01715 m

or

λ₁ = 1.7 𝐱 10⁻² m ------------Ans.1



For Lowest frequency sound, the above equation will become as

λ₂ = `\frac{v}{f₂}`     ∴{as v₁ =v₂ =v}

λ₂ = `\frac{343 ms⁻¹}{20 Hz }`  

∴{as f = `\frac{1}{T}` so, s⁻¹ = Hz}

λ₂ = `\frac{343 ms⁻¹}{20 s⁻¹ }` 

by simplifying we get

λ₂ = 17.15 m  ---------------Ans.2


Numerical Problems 11.9: A sound wave has a frequency of 2kHz and wavelength 35cm. How long will it take to travel 1.5km? Ans. (2.1 s)

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Given Data:


Frequency ( f ) = 2 KHz = 2 𝐱 10³ Hz

Wavelength ( λ ) = 35 cm = 0.35 m (conversion to SI unit)

Distance (S) = 1.5 Km = 1.5 𝐱 1000 m = 1500 m



To Find:


Time ( t ) = ? sec  



Solution:


To time (t), We have an equation S = v t , But we will have find first The speed of Sound ( v )from the given data be relation

 
v = f λ 

v = 2 𝐱 10³ Hz 𝐱 0.35 m

v = 700 m s⁻¹


Now 

S = v t

t = `\frac {S}{v}`

t = `\frac {1500 m}{700 m s⁻¹}`

by simplifying we get

t = 2.1 s  --------------Ans