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WAVE MOTION AND WAVES ON A STRING
EXERCISES:- Q-31 to Q-40
31. Two waves traveling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude?
ANSWER: Here given that amplitudes A₁ = A₂ =4 mm and δ = 90°. Since the frequencies and the wavelengths are similar, from the superposition principle the amplitude of the resultant wave is given as
A = √(A₁²+A₂²+2A₁A₂.cosδ)
→A = √(2A₁² + 2A₁².cos90°) =√(2A₁²)
→A = A₁√2 =4√2 mm
31. Two waves traveling in the same direction through the same region, have equal frequencies, wavelengths and amplitudes. If the amplitude of each wave is 4 mm and the phase difference between the waves is 90°, what is the resultant amplitude?
ANSWER: Here given that amplitudes A₁ = A₂ =4 mm and δ = 90°. Since the frequencies and the wavelengths are similar, from the superposition principle the amplitude of the resultant wave is given as
A = √(A₁²+A₂²+2A₁A₂.cosδ)
→A = √(2A₁² + 2A₁².cos90°) =√(2A₁²)
→A = A₁√2 =4√2 mm
ANSWER: Here given that amplitudes A₁ = A₂ =4 mm and δ = 90°. Since the frequencies and the wavelengths are similar, from the superposition principle the amplitude of the resultant wave is given as
A = √(A₁²+A₂²+2A₁A₂.cosδ)
→A = √(2A₁² + 2A₁².cos90°) =√(2A₁²)
→A = A₁√2 =4√2 mm
32. Figure (15-E7) shows two wave pulses at t = 0 traveling on a string in opposite directions with the same wave speed 50 cm/s. Sketch the shape of the string at t = 4 ms, 6 ms, 8 ms and 12 ms.
Figure for Q-32
ANSWER: Wave speed v = 50 cm/s =500 mm/s.
At t = 4 ms =0.004 s, the distance traveled by each of the pulses =500*0.004 mm =2 mm. So the pulses will come closer by 4 mm. The shape of the string will be as below.
Diagram for t = 4 ms
At t = 6 ms =0.006 s, the distance traveled by each of the pulses =500*0.006 mm =3 mm. So the pulses will come closer by 6 mm. The shape of the string will be as below.
Diagram for t = 6 ms
In the dotted region the displacements of string particles due to the two pulses are equal and opposite, hence net displacement is zero.
At t = 8 ms =0.008 s, the distance traveled by each of the pulses =500*0.008 mm =4 mm. So both the pulses will be between 6 mm and 10 mm. Since the pulses are inverted, the shape of the string at this overlap will be straight as below.
Diagram for t = 8 ms
At t = 12 ms =0.012 s, the distance traveled by each of the pulses =500*0.012 mm =6 mm. So the left pulse will be between 8 mm to 12 mm and the right pulse will be between 4 mm and 8 mm. Shape of the string will be as below.
Diagram for t = 12 ms
32. Figure (15-E7) shows two wave pulses at t = 0 traveling on a string in opposite directions with the same wave speed 50 cm/s. Sketch the shape of the string at t = 4 ms, 6 ms, 8 ms and 12 ms.
ANSWER: Wave speed v = 50 cm/s =500 mm/s.
At t = 4 ms =0.004 s, the distance traveled by each of the pulses =500*0.004 mm =2 mm. So the pulses will come closer by 4 mm. The shape of the string will be as below.
At t = 6 ms =0.006 s, the distance traveled by each of the pulses =500*0.006 mm =3 mm. So the pulses will come closer by 6 mm. The shape of the string will be as below.
At t = 12 ms =0.012 s, the distance traveled by each of the pulses =500*0.012 mm =6 mm. So the left pulse will be between 8 mm to 12 mm and the right pulse will be between 4 mm and 8 mm. Shape of the string will be as below.
 |
| Figure for Q-32 |
ANSWER: Wave speed v = 50 cm/s =500 mm/s.
At t = 4 ms =0.004 s, the distance traveled by each of the pulses =500*0.004 mm =2 mm. So the pulses will come closer by 4 mm. The shape of the string will be as below.
 |
| Diagram for t = 4 ms |
At t = 6 ms =0.006 s, the distance traveled by each of the pulses =500*0.006 mm =3 mm. So the pulses will come closer by 6 mm. The shape of the string will be as below.
 |
| Diagram for t = 6 ms |
In the dotted region the displacements of string particles due to the two pulses are equal and opposite, hence net displacement is zero.
At t = 8 ms =0.008 s, the distance traveled by each of the pulses =500*0.008 mm =4 mm. So both the pulses will be between 6 mm and 10 mm. Since the pulses are inverted, the shape of the string at this overlap will be straight as below.
 |
| Diagram for t = 8 ms |
At t = 12 ms =0.012 s, the distance traveled by each of the pulses =500*0.012 mm =6 mm. So the left pulse will be between 8 mm to 12 mm and the right pulse will be between 4 mm and 8 mm. Shape of the string will be as below.
 |
| Diagram for t = 12 ms |
33. Two waves each having a frequency of 100 Hz and a wavelength of 2.0 cm, are traveling in the same direction on a string. What is the phase difference the waves
(a) if the second wave was produced 0.015 s later than the first one at the same place,
(b) if the two waves were produced at the same instant but the first one was produced a distance 4.0 cm behind the second one?
(c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitude of the resultant waves in part (a) and (b)?
ANSWER: The frequency ν = 100 Hz. Wavelength 𝜆= 2.0 cm.
The wave speed v =𝜆ν = 2.0*100 =200 cm/s
(a) As in the diagram below, the OA describes the first wave that started a time t ago while the OB describes the second wave which has started just now. Clearly the phase difference ⲫ =⍵t.
So for t = 0.015 s,
The phase difference ⲫ = ⍵t =2πνt
→ⲫ =2π*100*0.015
→ⲫ = 3π
(b) The wave equation can be written y = A sin2π(t/T-x/𝜆)
Since both waves have been started at the same instant, at time t = 0, the wave equation at any distance x becomes
y = A sin(-2πx/𝜆) ................ (i)
Now consider a point on the string at a distance x-ẟx distance. Now y = A sin {-2π(x-δx)/𝜆} =A sin {-2πx/𝜆 +2πẟx/𝜆}
Comparing it (i) we get the phase difference ⲫ = 2πδx/𝜆
We have ẟx = 4 cm, 𝜆 = 2 cm
So, ⲫ = 2π*4/2 = 4π
(c) For the same frequency of waves, the resultant amplitude is given as
A = √(A₁²+A₂²+2A₁A₂.cosⲫ)
Here A₁ = A₂,
So A = √(2A₁²+2A₁².cosⲫ) =A₁√2*√(1+cosⲫ)
For part (a), ⲫ = 3π and A₁ =2 mm
Hence A = 2√2*√(1+cos3π) =2√2*√(1-1) = 0 i.e. zero.
For part (b), ⲫ = 4π
Hence A = 2√2*√(1+cos4π) =2√2*√2 =4.0 mm
33. Two waves each having a frequency of 100 Hz and a wavelength of 2.0 cm, are traveling in the same direction on a string. What is the phase difference the waves
(a) if the second wave was produced 0.015 s later than the first one at the same place,
(b) if the two waves were produced at the same instant but the first one was produced a distance 4.0 cm behind the second one?
(c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitude of the resultant waves in part (a) and (b)?
ANSWER: The frequency ν = 100 Hz. Wavelength 𝜆= 2.0 cm.
The wave speed v =𝜆ν = 2.0*100 =200 cm/s
(a) As in the diagram below, the OA describes the first wave that started a time t ago while the OB describes the second wave which has started just now. Clearly the phase difference ⲫ =⍵t.
So for t = 0.015 s,
The phase difference ⲫ = ⍵t =2πνt
→ⲫ =2π*100*0.015
→ⲫ = 3π
(b) The wave equation can be written y = A sin2π(t/T-x/𝜆)
Since both waves have been started at the same instant, at time t = 0, the wave equation at any distance x becomes
y = A sin(-2πx/𝜆) ................ (i)
Now consider a point on the string at a distance x-ẟx distance. Now y = A sin {-2π(x-δx)/𝜆} =A sin {-2πx/𝜆 +2πẟx/𝜆}
Comparing it (i) we get the phase difference ⲫ = 2πδx/𝜆
We have ẟx = 4 cm, 𝜆 = 2 cm
So, ⲫ = 2π*4/2 = 4π
(c) For the same frequency of waves, the resultant amplitude is given as
A = √(A₁²+A₂²+2A₁A₂.cosⲫ)
Here A₁ = A₂,
So A = √(2A₁²+2A₁².cosⲫ) =A₁√2*√(1+cosⲫ)
For part (a), ⲫ = 3π and A₁ =2 mm
Hence A = 2√2*√(1+cos3π) =2√2*√(1-1) = 0 i.e. zero.
For part (b), ⲫ = 4π
Hence A = 2√2*√(1+cos4π) =2√2*√2 =4.0 mm
(a) if the second wave was produced 0.015 s later than the first one at the same place,
(b) if the two waves were produced at the same instant but the first one was produced a distance 4.0 cm behind the second one?
(c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitude of the resultant waves in part (a) and (b)?
ANSWER: The frequency ν = 100 Hz. Wavelength 𝜆= 2.0 cm.
The wave speed v =𝜆ν = 2.0*100 =200 cm/s
(a) As in the diagram below, the OA describes the first wave that started a time t ago while the OB describes the second wave which has started just now. Clearly the phase difference ⲫ =⍵t.
The phase difference ⲫ = ⍵t =2πνt
→ⲫ =2π*100*0.015
→ⲫ = 3π
(b) The wave equation can be written y = A sin2π(t/T-x/𝜆)
Since both waves have been started at the same instant, at time t = 0, the wave equation at any distance x becomes
y = A sin(-2πx/𝜆) ................ (i)
Now consider a point on the string at a distance x-ẟx distance. Now y = A sin {-2π(x-δx)/𝜆} =A sin {-2πx/𝜆 +2πẟx/𝜆}
Comparing it (i) we get the phase difference ⲫ = 2πδx/𝜆
We have ẟx = 4 cm, 𝜆 = 2 cm
So, ⲫ = 2π*4/2 = 4π
(c) For the same frequency of waves, the resultant amplitude is given as
A = √(A₁²+A₂²+2A₁A₂.cosⲫ)
Here A₁ = A₂,
So A = √(2A₁²+2A₁².cosⲫ) =A₁√2*√(1+cosⲫ)
For part (a), ⲫ = 3π and A₁ =2 mm
Hence A = 2√2*√(1+cos3π) =2√2*√(1-1) = 0 i.e. zero.
For part (b), ⲫ = 4π
Hence A = 2√2*√(1+cos4π) =2√2*√2 =4.0 mm
34. If the speed of a transverse wave on a stretched string of length 1 m is 60 m/s, what is the fundamental frequency of vibration?
ANSWER: The length of the string, L = 1 m, wave speed V = 60 m/s. The frequency of a standing wave is given as ν = nV/2L.
For the fundamental frequency n = 1, so the fundamental frequency
=V/2L =60/(2*1) = 30 Hz
34. If the speed of a transverse wave on a stretched string of length 1 m is 60 m/s, what is the fundamental frequency of vibration?
ANSWER: The length of the string, L = 1 m, wave speed V = 60 m/s. The frequency of a standing wave is given as ν = nV/2L.
For the fundamental frequency n = 1, so the fundamental frequency
=V/2L =60/(2*1) = 30 Hz
35. A wire of length 2.0 m is stretched to a tension of 160 N. If the fundamental frequency of vibration is 100 Hz, Find its linear mass density.
ANSWER: The length L = 2.0 m, fundamental frequency ν =100 Hz, If V = speed of the wave,
ν = V/2L
→V =2νL =2*100*2 m/s =400 m/s.
But also V =√(F/µ), where F = tension and µ = linear mass density.
→µ = F/V² =160/400² kg/m =160*10³/400² g/m
→µ = 16/16 =1.00 g/m
35. A wire of length 2.0 m is stretched to a tension of 160 N. If the fundamental frequency of vibration is 100 Hz, Find its linear mass density.
ANSWER: The length L = 2.0 m, fundamental frequency ν =100 Hz, If V = speed of the wave,
ν = V/2L
→V =2νL =2*100*2 m/s =400 m/s.
But also V =√(F/µ), where F = tension and µ = linear mass density.
→µ = F/V² =160/400² kg/m =160*10³/400² g/m
→µ = 16/16 =1.00 g/m
ANSWER: The length L = 2.0 m, fundamental frequency ν =100 Hz, If V = speed of the wave,
ν = V/2L
→V =2νL =2*100*2 m/s =400 m/s.
But also V =√(F/µ), where F = tension and µ = linear mass density.
→µ = F/V² =160/400² kg/m =160*10³/400² g/m
→µ = 16/16 =1.00 g/m
36. A steel wire of mass 4.0 g and length 80 cm is fixed at two ends. The tension in the wire is 50 N. Find the frequency and the wavelength of the fourth harmonic of the fundamental.
ANSWER: The frequency of the fourth harmonic of the fundamental ν₄ =4V/2L. Here L = 80 cm = 0.80 m, µ = 4/(0.80*1000) kg/m = 0.005 kg/m, F = 50 N. Hence the wave speed
V = √(F/µ) =√(50/0.005) =√10000 =100 m/s.
Now ν₄ =4V/2L =4*100/(2*0.80) Hz =400/1.6 Hz = 250 Hz
Since V =νλ, The corresponding wavelength λ =V/ν
→λ = 100/250 m = 0.40 m =40 cm
36. A steel wire of mass 4.0 g and length 80 cm is fixed at two ends. The tension in the wire is 50 N. Find the frequency and the wavelength of the fourth harmonic of the fundamental.
ANSWER: The frequency of the fourth harmonic of the fundamental ν₄ =4V/2L. Here L = 80 cm = 0.80 m, µ = 4/(0.80*1000) kg/m = 0.005 kg/m, F = 50 N. Hence the wave speed
V = √(F/µ) =√(50/0.005) =√10000 =100 m/s.
Now ν₄ =4V/2L =4*100/(2*0.80) Hz =400/1.6 Hz = 250 Hz
Since V =νλ, The corresponding wavelength λ =V/ν
→λ = 100/250 m = 0.40 m =40 cm
ANSWER: The frequency of the fourth harmonic of the fundamental ν₄ =4V/2L. Here L = 80 cm = 0.80 m, µ = 4/(0.80*1000) kg/m = 0.005 kg/m, F = 50 N. Hence the wave speed
V = √(F/µ) =√(50/0.005) =√10000 =100 m/s.
Now ν₄ =4V/2L =4*100/(2*0.80) Hz =400/1.6 Hz = 250 Hz
Since V =νλ, The corresponding wavelength λ =V/ν
→λ = 100/250 m = 0.40 m =40 cm
37. A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the "Middle C" (ν = 261.63 Hz). Find the tension in the wire.
ANSWER: The fundamental frequency ν₀ =V/2L.
Given L = 90 cm. ν₀ = 261.63
→V =2ν₀L =2*261.63*(90/100) =470.93 m/s
But V =√(F/µ)
→F = µV² =(6*100/90*1000)*470.93² ≈1480 N
37. A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the "Middle C" (ν = 261.63 Hz). Find the tension in the wire.
ANSWER: The fundamental frequency ν₀ =V/2L.
Given L = 90 cm. ν₀ = 261.63
→V =2ν₀L =2*261.63*(90/100) =470.93 m/s
But V =√(F/µ)
→F = µV² =(6*100/90*1000)*470.93² ≈1480 N
ANSWER: The fundamental frequency ν₀ =V/2L.
Given L = 90 cm. ν₀ = 261.63
→V =2ν₀L =2*261.63*(90/100) =470.93 m/s
But V =√(F/µ)
→F = µV² =(6*100/90*1000)*470.93² ≈1480 N
38. A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire?
ANSWER: The frequency of the second harmonic ν₂ =256 Hz.
L = 1.50 m. If the speed of the wave = V, then ν₂ = 2V/2L =V/L
→V =ν₂L =256*1.5 m/s =384 m/s
38. A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire?
ANSWER: The frequency of the second harmonic ν₂ =256 Hz.
L = 1.50 m. If the speed of the wave = V, then ν₂ = 2V/2L =V/L
→V =ν₂L =256*1.5 m/s =384 m/s
ANSWER: The frequency of the second harmonic ν₂ =256 Hz.
L = 1.50 m. If the speed of the wave = V, then ν₂ = 2V/2L =V/L
→V =ν₂L =256*1.5 m/s =384 m/s
39. The length of the wire shown in figure (15-E8) between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
Figure for Q-39
ANSWER: The vibration in two loops leaving the middle point of the wire between the pulleys at rest means it is vibrating in second harmonic. Thus ν₂ =2V/2L =V/L = (1/L)√(F/µ)
Here L = 1.5 m, mass m =12 g = 0.012 kg,
µ =m/L =0.012/1.5 kg/m =0.008 kg/m
F =9g N =9*9.8 N =88.2 N {Taking g = 9.8 m/s²}
Now ν₂ = (1/1.5)√(88.2/0.008) Hz
=(1/1.5)√11025 Hz =105/1.5 Hz = 70 Hz
39. The length of the wire shown in figure (15-E8) between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
ANSWER: The vibration in two loops leaving the middle point of the wire between the pulleys at rest means it is vibrating in second harmonic. Thus ν₂ =2V/2L =V/L = (1/L)√(F/µ)
Here L = 1.5 m, mass m =12 g = 0.012 kg,
µ =m/L =0.012/1.5 kg/m =0.008 kg/m
F =9g N =9*9.8 N =88.2 N {Taking g = 9.8 m/s²}
Now ν₂ = (1/1.5)√(88.2/0.008) Hz
=(1/1.5)√11025 Hz =105/1.5 Hz = 70 Hz
 |
| Figure for Q-39 |
ANSWER: The vibration in two loops leaving the middle point of the wire between the pulleys at rest means it is vibrating in second harmonic. Thus ν₂ =2V/2L =V/L = (1/L)√(F/µ)
Here L = 1.5 m, mass m =12 g = 0.012 kg,
µ =m/L =0.012/1.5 kg/m =0.008 kg/m
F =9g N =9*9.8 N =88.2 N {Taking g = 9.8 m/s²}
Now ν₂ = (1/1.5)√(88.2/0.008) Hz
=(1/1.5)√11025 Hz =105/1.5 Hz = 70 Hz
40. A one meter long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops?
ANSWER: The vibration in four loops means the string is vibrating in fourth harmonic. The frequency of the fourth harmonic is given as ν₄ =4V/2L =2V/L, where V = wave speed and L = length of the string = 1.0 m. Given ν₄ = 128 Hz.
Hence the wave speed V = ν₄L/2 =128*1/2 =64 m/s.
Mass of the string m =40 g = 0.040 kg.
Linear mass density µ = m/L =0.04/1 =0.04 kg/m
Since V =√(F/µ)
→F = µV² =0.04*64² Hz ≈ 164 Hz
40. A one meter long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops?
ANSWER: The vibration in four loops means the string is vibrating in fourth harmonic. The frequency of the fourth harmonic is given as ν₄ =4V/2L =2V/L, where V = wave speed and L = length of the string = 1.0 m. Given ν₄ = 128 Hz.
Hence the wave speed V = ν₄L/2 =128*1/2 =64 m/s.
Mass of the string m =40 g = 0.040 kg.
Linear mass density µ = m/L =0.04/1 =0.04 kg/m
Since V =√(F/µ)
→F = µV² =0.04*64² Hz ≈ 164 Hz
Hence the wave speed V = ν₄L/2 =128*1/2 =64 m/s.
Mass of the string m =40 g = 0.040 kg.
Linear mass density µ = m/L =0.04/1 =0.04 kg/m
Since V =√(F/µ)
→F = µV² =0.04*64² Hz ≈ 164 Hz
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Links to the Chapters
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Links to the Chapters
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
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- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
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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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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