Friday, December 28, 2018

Solutions to Problems on "SOUND WAVES" - H C Verma's Concepts of Physics, Part-I, Chapter-16, QUESTIONS FOR SHORT ANSWER

My Channel on YouTube  →  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

SOUND WAVES

QUESTIONS FOR SHORT ANSWER

1. If you are walking on the moon, can you hear the sound of stones cracking behind you? Can you hear the sound of your own footsteps?

ANSWER: Since there is no air on the moon, there is the absence of a medium in which the cracking stones would set vibrations. The vibrations of the stones can not propagate in absence of a medium. So the sound of the stones cracking behind cannot be heard. In fact, no sound is produced. But this answer is on the assumption that our ears are open to the environment like on the earth which is not correct. In the absence of the air, there is no atmospheric pressure of which we are accustomed to. It will damage our body and ears. So we need space suits which isolates the whole body from the surrounding.
         The sound of own footsteps can be heard because the vibrations produced can travel through our body to the ears.  


2. Can you hear your own words if you are standing in a perfect vacuum? Can you hear your friend in the same conditions?

ANSWER: When we speak the vocal cords in our throat vibrate. These vibrations reach our ears internally. So even in a perfect vacuum, we can hear our own words.   
        But in the same conditions, we cannot hear our friend due to the absence of a medium which allows propagating the vibrations from one point to the other. 


3. A vertical rod is hit at one end. What kind of wave propagates in the rod if (a) the hit is made vertically (b) the hit is made horizontally?

ANSWER: (a) The vertical hit will set the particles at that end to vibrate longitudinally, This longitudinal disturbance propagates as a longitudinal wave in the rod.
              (b) The horizontal hit will set the particles at that end to vibrate along the perpendicular to the axis of the rod. So the disturbance will propagate as a transverse wave in the rod.  


4. Two loudspeakers are arranged facing each other at some distance. Will a person standing behind one of the loudspeakers clearly hear the sound of the other loudspeaker or the clarity will be seriously damaged because of the "collision" of the two sounds in between? 

ANSWER: The loudspeakers have very high intensity in the front direction but not so high in the backward direction. In the given condition let us assume that both loudspeakers are connected to the same source. The person behind the first loudspeaker will hear comparatively very low-intensity sound from it than the other one if the distance between the loudspeakers is small. Even if their sound interferes either constructively or destructively at the ears of the person, he will hear the sound clearly. But if the distance between the loudspeakers is large, the intensity of the other loudspeaker will be comparable to the first one, also due to the path difference both sources will have a phase difference and the clarity of the sound will be seriously affected.


5. The voice of a person, who has inhaled helium, has a remarkably high pitch. Explain on the basis of the resonant vibration of the vocal cord filled with air and with helium. 

ANSWER: The resonant frequency of the vocal cord in the voice box is directly proportional to the speed of the sound. The voice box of a person inhaling helium has helium-filled in it. The speed of sound in helium is about three times more than in the air. Hence the resonant frequency is higher in this case. Therefore his voice will be of high pitch. 


6. Draw a diagram to show the standing pressure wave and standing displacement wave for the 3rd overtone mode of vibration of an open organ pipe.

ANSWER: Frequencies for a standing wave in an open organ pipe is given by, ν = nV/2L, where n = 1, 2, 3, .....
n = 1 is for the fundamental mode of vibration. For the 3rd overtone mode of vibration n = 4, and the corresponding frequency
ν = 4V/2L. It will have four pressure antinodes. Corresponding to these antinodes there will be four displacement nodes. The diagram is the following:-
Diagram for Q - 6


7. Two tuning forks vibrate with the same amplitude but the frequency of the first is double the frequency of the second. Which fork produces more intense sound in air? 

ANSWER: The amplitudes of both tuning forks are same hence both will produce same displacement amplitudes in the sound wave in the air say s₀. Corresponding pressure amplitudes
p₀ = B⍵s₀/V =B*2πν*s₀/V =(2πBs₀/V)ν
Since the intensity of sound is proportional to the square of the pressure amplitude p₀, but p₀ is proportional to the frequency ν hence the intensity of the sound is also proportional to the square of the frequency. Hence the first tuning fork having double frequency will produce more intense sound in air.


8. In discussing the Doppler effect, we use the word "apparent frequency". Does it mean that the frequency of the sound is still that of the source and it is some physiological phenomenon in the listener's ear that gives rise to the Doppler effect? Think for the observer approaching the source and for the source approaching the observer.

ANSWER: The frequency of the sound remains the same that of the source. The apparent frequency refers to the frequency perceived by the observer due to the relative motion of the observer and the source.
   When the observer approaches the source, the time between two consecutive points of the same phase decreases, so he perceives the sound as having a higher frequency than when standing.
When the source approaches the observer again due to the relative motion the second consecutive point of the same phase is produced nearer to the observer, so the two consecutive points of the same phase reach observer in a shorter time than both were not moving. So again the observer perceives the sound as having a higher frequency. This phenomenon is named as Doppler effect.

===<<<O>>>=== 

Links to the Chapters

CHAPTER- 12 - Simple Harmonic Motion



EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)



CHAPTER- 11 - Gravitation



EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion


Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"


Monday, December 17, 2018

Solutions to Problems on "WAVE MOTION AND WAVES ON A STRING" - H C Verma's Concepts of Physics, Part-I, Chapter-15, EXERCISES Q_51 to Q_57

My Channel on YouTube  →  SimplePhysics with KK

For links to 

other chapters - See bottom of the page

Or click here → kktutor.blogspot.com

WAVE MOTION AND WAVES ON A STRING

EXERCISES:- Q-41 to Q-50

51. A 2 m long string fixed at both ends is set into vibrations in its first overtone. The wave speed on the string is 200 m/s and the amplitude is 0.5 cm.
(a) Find the wavelength and the frequency.
(b) Write the equation giving the displacement of different points as a function of time. Choose the x-axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.

ANSWER: (a) Wave speed V = 200 m/s. Amplitude = 0.5 cm. The length of the string L = 2 m. The vibration is in first overtone i.e. second harmonic. Hence the frequency
ν = 2V/2L =V/L = 200/2 Hz = 100 Hz.
The wavelength 𝜆 = V/ν =200/100 m = 2.0 m.

(b) The equation of a standing wave is given as
y = 2A sin kx cos ⍵t
Here ⍵ = 2πν =2π*100 s⁻¹ =200π s⁻¹
k = ⍵/V = 200π/200 m⁻¹ = π m⁻¹
Now, y = 2A sin[(π m⁻¹)x] cos [(200π s⁻¹)t]
Given condition is, at t = 0 and x = 50cm = 0.5 m, y = maximum displacement i.e. amplitude = 0.5 cm; putting it in the equation,
0.5 cm = 2A sin(π/2) cos 0 =2A
Thus 2A = 0.5 cm. Hence the required equation is
y = (0.5 cm) sin[(π m⁻¹)x] cos [(200π s⁻¹)t].

52. The equation for the vibration of a string, fixed at both ends vibrating in its third harmonic, is given by
y = (0.04 cm) sin[(0.314 cm⁻¹)x] cos [(600π s⁻¹)t].
(a) What is the frequency of vibration?
(b) What are the positions of the node?
(c) What is the length of the string?
(d) What are the wavelength and the speed of two traveling waves that can interfere to give this vibration?

ANSWER: (a) Comparing the given equation with the general equation of the standing wave
y = 2A sin kx cos ⍵t
⍵ = 600π s⁻¹
→2πν = 600π
→ν = 600/2 Hz =300 Hz

(b) Comparing the given equation with the general equation of the standing wave
y = 2A sin kx cos ⍵t
k = 0.314 cm⁻¹
But k = ⍵/V = 2πν/ν𝜆 =2π/𝜆
→𝜆 = 2π/k =2π/0.314 =20 cm
In the third harmonic vibrations, there will be three loops with four nodes equally spaced at 𝜆/2 distance i.e. 20/2 cm = 10 cm. The fixed ends are essentially the nodes.
Hence the nodes are 
at x = 0, x = 10 cm, x = 20 cm and x = 30 cm. 
The diagram for Q - 52


(c) The distance between the first and the last nodes will be the length of the string. Here first node is x = 0 and the last node is x = 30 cm. Hence the length of the string is = 30 cm.

Alternately,
Since the vibration is in third harmonic, the frequency
ν = 3V/2L =3⍵/2kL, {Since k =⍵/V}
→L = 1.5⍵/kν
→L = 1.5*600π/(0.314*300) 
→L =1.5*2*3.14/0.314 =3*10 =30 cm
Hence the length of the string is 30 cm.

(d) As we have calculated the wavelength of the interfering waves in (b),
𝜆 = 20 cm
The speed of the wave V = ⍵/k = 600π/0.314 cm/s
=6000 cm/s =60 m/s


53. The equation of a standing wave, produced on a string fixed at both ends, is
y = (0.04 cm) sin[(0.314 cm⁻¹)x] cos [(600π s⁻¹)t].
What could be the smallest length of the string?

ANSWER: From the equation ⍵ = 600π s⁻¹, k =0.314 cm⁻¹
But k = 2π/𝜆,
→𝜆 = 2π/k =2*3.14/0.314 = 20 cm
For the nth harmonic the frequency
ν = nV/2L
→L =nV/2ν = n𝜆/2 = n*20/2 cm =10*n cm
The smallest length of the string will be for fundamental harmonic for n = 1,
So L = 10*1 =10 cm


54. A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of the cross-section of wire is 1.0 mm², Find its Young's Modulus.

ANSWER: The elongation of the wire = 40.05 - 40 cm =0.05 cm =0.05/100 m = 5 x 10⁻⁴ m. 
Strain ε =elongation/length =5x10⁻⁴/0.4 =1.25 x 10⁻³
To find Young's modulus we need to find the stress in the wire. Let the tension in the wire be F. Fundamental frequency ν = 220 Hz. 
Hence V/2L =ν, {where V = wave speed}
→V =2νL =2*220*0.4 m/s = 176 m/s
The linear mass density
µ = 3.2*100/(40*1000) kg/m = 0.008 kg/m
Wave speed V =√(F/µ)
→F = µV² =0.008*176² =247.8 N
The area of cross-section =1 mm² = 1 x 10⁻⁶ m²
Stress in the wire σ = 247.8/(1 x 10⁻⁶) N/m² =247.8 x 10⁶ N/m²
Young's Modulus Y =σ/ε = 247.8 x 10⁶/(1.25x10⁻³) N/m²
→Y = 198 x 10⁹ N/m²
Y = 1.98 x 10¹¹ N/m²


55. Figure (15-E11) shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block.
The figure for Q - 55


ANSWER: Let the length of the string = L
The frequency of the tenth harmonic ν = 10V/2L
where V is the wave speed.
The frequency of the eleventh harmonic is also the same but the wave speed is changed due to the change in the wire tension. Now 
ν = 11V'/2L
where V' is the wave speed in the second case.
So, 10V/2L = 11V'/2L
→V = 1.1V'
→√(F/µ) = 1.1√(F'/µ)
→F' = F/1.21
Let the density of the block = ρ and volume = v.
Mass of the block =ρv
The tension in the string in the first case = weight of the block in the air
=F = ρvg
The loss of the weight of the block = weight of the same volume of the water = ρ'vg
where ρ' = density of water.
Hence F - F' = ρ'vg
→F-F/1.21 =ρ'vg
→0.21F =1.21ρ'vg
→F = 5.8 ρ'vg
ρvg = 5.8 ρ'vg
→ρ = 5.8 ρ' =5.8 x 10³ kg/m³
(Since the density of water ρ' = 1000 kg/m³)


56. A 2.00 m long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N.
(a) Find the frequencies of the fundamental and the first two overtones. 
(b) Find the wavelength of in the fundamental and the first two overtones. 

ANSWER: (a) For the given condition of end fixity, the frequency of nth harmonic of the fundamental ν = (n+½)V/2L
Where V = wave speed and n = 0, 1, 2 .... 
Linear mass density µ = 0.08/2 kg/m =0.04 kg/m
V =√(F/µ) =√(256/0.04) =√6400 m/s =80 m/s
Hence the frequency ν = (n+½)*80/(2*2) =20(n+½) Hz
The fundamental frequency for n = 0 is =20*½ = 10 Hz
The frequency of first overtone is for n = 1 which is =20(1+½) Hz
=20*3/2 Hz =30 Hz
The frequency of second overtone is for n = 2 which is =20(2+½) Hz
=20*5/2 Hz = 50 Hz

(b) Since ν = (n+½)V/2L 
→ν = (n+½)*ν𝜆/2L
→𝜆 = 2L/(n+½)
Hence wavelength of the fundamental vibration for n = 0 is
𝜆 = 2L/½ =4L =4*2 m = 8.00 m

The wavelength for the first overtone is for n = 1
𝜆₁ = 2L/(3/2) =4L/3 =4*2/3 = 2.67 m

The wavelength for the second overtone is for n = 2

𝜆₂ = 2L/(5/2) =4L/5 =4*2/5 = 1.60 m


57. A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure (15-E12). The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string resonate?
The figure for Q - 57


ANSWER: The heavy string has one end fixed and the other end free to move in the transverse direction. In this case for the standing wave, the frequency of vibration is given as 
ν =(n+½)V/2L
Minimum frequency is for n = 0,
i.e. V/4L = 120 Hz, {Given}
→V/L = 480

In the second case when the joint is on the pulley, the end condition changes and both ends become fixed. Now the frequency for the standing wave = nV/2L.
Minimum frequency is for n = 1, hence 
Minimum frequency = V/2L =½(V/L) =½*480 Hz =240 Hz

===<<<O>>>=== 

Links to the Chapters

CHAPTER- 12 - Simple Harmonic Motion



EXERCISES- Q1 TO Q10

EXERCISES- Q11 TO Q20

EXERCISES- Q21 TO Q30

EXERCISES- Q31 TO Q40

EXERCISES- Q41 TO Q50

EXERCISES- Q51 TO Q58 (2-Extra Questions)



CHAPTER- 11 - Gravitation



EXERCISES -Q 31 TO 39

CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Vector related Problems"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"