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WAVE MOTION AND WAVES ON A STRING
QUESTIONS FOR SHORT ANSWER
1. You are walking along a seashore and a mild wind is blowing. Is the motion of air a wave motion.
ANSWER: The motion of air is not a wave motion because in a wave motion the particles of the medium do not move from one place to another but vibrate at its place.
2. The radio and TV programmes, telecast at the studio, reach our antenna by wave motion. Is it a mechanical wave or non-mechanical?
ANSWER: It is a non-mechanical wave because it does not require a medium. These waves travel from ground to satellites and then relayed back to our antenna. There is no medium in the space near the satellite.
3. A wave is represented by an equation y = c₁ sin (c₂.x+c₃.t). In which direction is the wave going? Assume that c₁, c₂ and c₃ are all positive.
ANSWER: y = c₁ sin (c₂.x+c₃.t)
→y = c₁ sin {c₃.(t + c₂.x/c₃)}
→y = c₁ sin [c₃.{t + x/(c₃/c₂)}]
→y = c₁ sin [c₃.(t + x/c)] = ƒ(t + x/c), where c = c₃/c₂.
So the wave is travelling in the negative x-direction.
4. Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π.
ANSWER: The wave equation can also be written as
y = A.sin2π(t/T-x/λ)
where y is the transverse displacement of the particle at a distance x at time t and A is the amplitude of the particle.
The speed of the particle can be found out by partial differentiation of y with respect to t.
∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→V = {2πA.cos 2π(t/T-x/λ)}/T
For V to be maximum cos2π(t/T-x/λ) = 1
So Vₘₐₓ = 2πA/T
if A =λ/2π
Vₘₐₓ = λ/T = v, where v is the wave speed.
But if A < λ/2π, then Vₘₐₓ < λ/T
→Vₘₐₓ < v.
So, if the amplitude is less than the wavelength divided by 2π, the particle speed can never be more than the wave speed.
5. Two wave pulses identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At an instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of the two pulses?
ANSWER: When the pulses reach the middle, though the string becomes completely straight, the particles of the string there do not stop but have speed. Due to the speed, the pulses again emerge out after a few moments. So the energy of the two pulses at the middle is in the form of kinetic energy of the particles.
6. Show that for a wave traveling on a string
yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ
where the symbols have usual meanings. Can we use Componendo and Dividendo taught in algebra to write
(yₘₐₓ+vₘₐₓ)/(yₘₐₓ-vₘₐₓ) = (vₘₐₓ+aₘₐₓ)/(vₘₐₓ-aₘₐₓ)?
ANSWER: y = A.sin2π(t/T-x/λ) ....... (i)
yₘₐₓ = A. {Because maximum value of sine = 1}
v = ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→vₘₐₓ = 2πA/T. {} Because also maximum value of cosine =1}
a = ∂²y/∂t² = A.sin2π(t/T-x/λ)*4π²/T²
→aₘₐₓ = 4π²A/T²
Now yₘₐₓ/vₘₐₓ =A/(2πA/T) = T/2π
and vₘₐₓ/aₘₐₓ = (2πA/T)/(4π²A/T²) =T/2π
Hence, yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ.
We cannot use Componendo and Dividendo because addition and subtraction are allowed only between quantities having same unit or dimension. Here yₘₐₓ, vₘₐₓ and aₘₐₓ all have different units and dimensions.
7. What is the smallest positive phase constant which is equivalent to 7.5π?
ANSWER: The displacement of a particle on the string is given as
y = A.sin[⍵(t-x/v) + ⲫ]
where ⲫ is the phase constant.
Since a value of the phase constant will be unique for one oscillation and it will be repeated at each 2π angle. So the smallest positive value of ⲫ equivalent to 7.5π will be = 7.5π-2nπ by putting n = 3 i.e., 7.5π-6π = 1.5π
8. A string clamped at both ends vibrates in its fundamental mode. Is there any position (except the ends) on the string which can be touched without disturbing the motion? What if the string vibrates in its first overtone?
ANSWER: No, because in the fundamental mode all points except the ends are vibrating with different amplitudes. Touching any point other than the ends will disturb the motion. When the string vibrates in the first overtone there is a node also at the midpoint which is stationary. This midpoint can be touched without disturbing the motion.
Figure for Q - 8
Or click here → kktutor.blogspot.com
WAVE MOTION AND WAVES ON A STRING
QUESTIONS FOR SHORT ANSWER
1. You are walking along a seashore and a mild wind is blowing. Is the motion of air a wave motion.
ANSWER: The motion of air is not a wave motion because in a wave motion the particles of the medium do not move from one place to another but vibrate at its place.
2. The radio and TV programmes, telecast at the studio, reach our antenna by wave motion. Is it a mechanical wave or non-mechanical?
ANSWER: It is a non-mechanical wave because it does not require a medium. These waves travel from ground to satellites and then relayed back to our antenna. There is no medium in the space near the satellite.
3. A wave is represented by an equation y = c₁ sin (c₂.x+c₃.t). In which direction is the wave going? Assume that c₁, c₂ and c₃ are all positive.
ANSWER: y = c₁ sin (c₂.x+c₃.t)
→y = c₁ sin {c₃.(t + c₂.x/c₃)}
→y = c₁ sin [c₃.{t + x/(c₃/c₂)}]
→y = c₁ sin [c₃.(t + x/c)] = ƒ(t + x/c), where c = c₃/c₂.
So the wave is travelling in the negative x-direction.
→y = c₁ sin {c₃.(t + c₂.x/c₃)}
→y = c₁ sin [c₃.{t + x/(c₃/c₂)}]
→y = c₁ sin [c₃.(t + x/c)] = ƒ(t + x/c), where c = c₃/c₂.
So the wave is travelling in the negative x-direction.
4. Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π.
ANSWER: The wave equation can also be written as
y = A.sin2π(t/T-x/λ)
where y is the transverse displacement of the particle at a distance x at time t and A is the amplitude of the particle.
The speed of the particle can be found out by partial differentiation of y with respect to t.
∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→V = {2πA.cos 2π(t/T-x/λ)}/T
For V to be maximum cos2π(t/T-x/λ) = 1
So Vₘₐₓ = 2πA/T
if A =λ/2π
Vₘₐₓ = λ/T = v, where v is the wave speed.
But if A < λ/2π, then Vₘₐₓ < λ/T
→Vₘₐₓ < v.
So, if the amplitude is less than the wavelength divided by 2π, the particle speed can never be more than the wave speed.
y = A.sin2π(t/T-x/λ)
where y is the transverse displacement of the particle at a distance x at time t and A is the amplitude of the particle.
The speed of the particle can be found out by partial differentiation of y with respect to t.
∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→V = {2πA.cos 2π(t/T-x/λ)}/T
For V to be maximum cos2π(t/T-x/λ) = 1
So Vₘₐₓ = 2πA/T
if A =λ/2π
Vₘₐₓ = λ/T = v, where v is the wave speed.
But if A < λ/2π, then Vₘₐₓ < λ/T
→Vₘₐₓ < v.
So, if the amplitude is less than the wavelength divided by 2π, the particle speed can never be more than the wave speed.
5. Two wave pulses identical in shape but inverted with respect to each other are produced at the two ends of a stretched string. At an instant when the pulses reach the middle, the string becomes completely straight. What happens to the energy of the two pulses?
ANSWER: When the pulses reach the middle, though the string becomes completely straight, the particles of the string there do not stop but have speed. Due to the speed, the pulses again emerge out after a few moments. So the energy of the two pulses at the middle is in the form of kinetic energy of the particles.
6. Show that for a wave traveling on a string
yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ
where the symbols have usual meanings. Can we use Componendo and Dividendo taught in algebra to write
(yₘₐₓ+vₘₐₓ)/(yₘₐₓ-vₘₐₓ) = (vₘₐₓ+aₘₐₓ)/(vₘₐₓ-aₘₐₓ)?
ANSWER: y = A.sin2π(t/T-x/λ) ....... (i)
yₘₐₓ = A. {Because maximum value of sine = 1}
v = ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→vₘₐₓ = 2πA/T. {} Because also maximum value of cosine =1}
a = ∂²y/∂t² = A.sin2π(t/T-x/λ)*4π²/T²
→aₘₐₓ = 4π²A/T²
Now yₘₐₓ/vₘₐₓ =A/(2πA/T) = T/2π
and vₘₐₓ/aₘₐₓ = (2πA/T)/(4π²A/T²) =T/2π
Hence, yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ.
We cannot use Componendo and Dividendo because addition and subtraction are allowed only between quantities having same unit or dimension. Here yₘₐₓ, vₘₐₓ and aₘₐₓ all have different units and dimensions.
yₘₐₓ = A. {Because maximum value of sine = 1}
v = ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T
→vₘₐₓ = 2πA/T. {} Because also maximum value of cosine =1}
a = ∂²y/∂t² = A.sin2π(t/T-x/λ)*4π²/T²
→aₘₐₓ = 4π²A/T²
Now yₘₐₓ/vₘₐₓ =A/(2πA/T) = T/2π
and vₘₐₓ/aₘₐₓ = (2πA/T)/(4π²A/T²) =T/2π
Hence, yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ.
We cannot use Componendo and Dividendo because addition and subtraction are allowed only between quantities having same unit or dimension. Here yₘₐₓ, vₘₐₓ and aₘₐₓ all have different units and dimensions.
7. What is the smallest positive phase constant which is equivalent to 7.5π?
ANSWER: The displacement of a particle on the string is given as
y = A.sin[⍵(t-x/v) + ⲫ]
where ⲫ is the phase constant.
Since a value of the phase constant will be unique for one oscillation and it will be repeated at each 2π angle. So the smallest positive value of ⲫ equivalent to 7.5π will be = 7.5π-2nπ by putting n = 3 i.e., 7.5π-6π = 1.5π
8. A string clamped at both ends vibrates in its fundamental mode. Is there any position (except the ends) on the string which can be touched without disturbing the motion? What if the string vibrates in its first overtone?
ANSWER: No, because in the fundamental mode all points except the ends are vibrating with different amplitudes. Touching any point other than the ends will disturb the motion. When the string vibrates in the first overtone there is a node also at the midpoint which is stationary. This midpoint can be touched without disturbing the motion.
 |
| Figure for Q - 8 |
===<<<O>>>===
Links to the Chapters
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Links to the Chapters
ALL LINKS
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
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
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CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-II
EXERCISES- Q-1 TO Q-10
EXERCISES- Q-11 TO Q-20
EXERCISES- Q-21 TO Q-32
CHAPTER- 13 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
OBJECTIVE-II
EXERCISES Q-1 TO Q-10
EXERCISES- Q11 TO Q20
EXERCISES Q-21 TO Q30
EXERCISES Q-31 TO Q35
CHAPTER- 12 - Simple Harmonic Motion
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
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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For more practice on problems on friction solve these- "New Questions on Friction".
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Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
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Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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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
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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"
CHAPTER- 2 - "Vector related Problems"
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