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WAVE MOTION AND WAVES ON A STRING
EXERCISES:- Q-41 to Q-50
41. A wire fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz.
(a) What could be the maximum value of the fundamental frequency?
(b) If transverse waves can travel on this string at speed of 40 m/s, what is its length?
ANSWER: (a) These resonant frequencies are some nth harmonic of the fundamental. So 240 Hz and 320 Hz are some multiples of the same fundamental frequency. The maximum frequency of the fundamental will be the HCF of these two frequencies.
HCF of 240 and 320 is 80. Hence the maximum value of fundamental frequency is 80 Hz.
(b) The speed of the wave V = 40 m/s. For the fundamental frequency ν = V/2L.
→L = V/2ν =40/(2*80) m =0.25 m = 25 cm.
42. A string fixed at both ends vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency this separation is reduced to 1.6 cm. Find the length of the string.
ANSWER: Let for the nth resonant mode the separation between consecutive node is 2.0 cm. Hence the length of the string = 2n cm.
For the next higher i.e. (n+1)th resonant mode the separation is 1.6 cm. From this the length of the string = 1.6*(n+1) cm. These two expressions for the length of the string will be equal.
→2n = 1.6(n+1) =1.6n + 1.6
→0.4n = 1.6
→n =1.6/0.4 = 4
Hence the length of the string = 2n cm = 2*4 cm = 8.0 cm
43. A 660 Hz tuning fork sets up a vibration in a string clamped at both ends. The wave speed for a transverse wave on the string is 220 m/s and the string vibrates in three loops.
(a) Find the length of the string.
(b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.
ANSWER: (a) Here the wave speed V = 220 m/s. Resonant frequency ν = 660 Hz. Since the string vibrates in three loops, it is vibrating in third harmonic. i.e. n = 3. Length of the string =?
ν = nV/2L
→660 = 3*220/2L
→L = 660/(2*660) =1/2 = 0.50 m =50 cm.
(b) The equation of a wave on a string fixed at both ends is given by
y =2A sin kx cos ⍵t
Where A is the amplitude of a wave on the wire. Here,
⍵ = 2πν =2π*660 s⁻¹ =1320π s⁻¹
k =2π/λ =2πν/V, Since λ = V/ν
→k = 2π*660/220 m⁻¹ =6π m⁻¹ =0.06π cm⁻¹
The standing wave is a result of interference of two waves in opposite direction having the same amplitude A. When the standing wave is formed on the string the maximum amplitude is the sum of the amplitudes of both the waves. The sum of amplitudes of both the waves = 2A (Twice the amplitude) = Maximum amplitude =0.50 cm
Now the wave equation is
y = (0.5 cm) sin [(0.06π cm⁻¹)x]*cos[(1320π s⁻¹)t]
44. A particular guitar wire is 30 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes?
ANSWER: Here ν = 196 Hz. Since no finger is placed its vibrating length L = 30 cm. The frequency is inversely proportional to the length of the string. Hence if other things are constant,
ν/ν' = L'/L
→L' = νL/ν'
For ν' = 220 Hz
The distance from the end where the finger be placed for this frequency = 196*30/220 cm = 26.7 cm
For ν' = 247 Hz
The distance from the end where the finger be placed for this frequency = 196*30/247 cm = 23.8 cm
For ν' = 262 Hz
The distance from the end where the finger be placed for this frequency = 196*30/262 cm = 22.4 cm
For ν' = 294 Hz
The distance from the end where the finger be placed for this frequency = 196*30/294 cm = 20.0 cm
45. A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear a sound of the maximum frequency of 14 kHz. What is the highest harmonic that can be played on this string which is audible to the person?
ANSWER: Here ν₀ = 200 Hz
→ V/2L = 200, {L = length of the string, V = wave speed}
Let the highest harmonic that can be played on this string and audible o the person = n. The frequency of this harmonic = νₙ
νₙ ⩽14 kHz
→nV/2L ⩽ 14000 Hz
→n*(V/2L) ⩽ 14000 Hz
→n*220 ⩽ 14000 Hz
→n ⩽ 14000/200
→n ⩽ 70
So the highest harmonic audible to the person = 70.
46. Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string.
(b) Which harmonics of the fundamental are the given frequencies?
(c) Which overtones are these frequencies?
(d) If the length of the string is 80 cm, what would be the speed of a transverse wave on the string?
ANSWER: (a) The highest possible fundamental frequency of vibration of the string will be the HCF of these three frequencies.
90 = 2*3*3*5
150 = 2*3*5*5
210 =2*3*5*7
HCF of 90, 150 and 210 Hz is =2*3*5 =30 Hz
(b) Since the higher harmonics are multiples of the fundamental harmonic,
90 Hz is the 90/30 = 3rd harmonic
150 Hz is the 150/30 =5th harmonic
210 Hz is the 210/30 =7th harmonic
(c) Since the nth harmonic is the (n-1)th overtone
3rd harmonic = 2nd overtone
5th harmonic = 4th overtone
7th harmonic = 6th overtone
(d) Given L = 80 cm =0.80 m, V = ?
For the fundamental frequency ν₀ = 30 Hz,
→V/2L = 30
→V = 30*2*0.80 = 48 m/s
47. Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2:1, the radii are in the ratio 3:1 and the densities are in the ratio 1:2. Find the ratio of their fundamental frequencies.
ANSWER: Let the tension in the wires = 2F and F and the radius = 3r and r. If the densities are ρ and 2ρ,
The linear mass density of first wire µ = π(3r)²*1*ρ =9πρr²
The linear mass density of the second wire µ' = πr²*2ρ =2πr²ρ
The fundamental frequency of the first wire ν₀ =V/2L =(1/2L)*√(2F/µ)
The fundamental frequency of the second wire ν₀' = (1/2L)*√(F/µ')
Hence the ratio of the fundamental frequencies
ν₀/ν₀' = √(2/µ)/√(1/µ') =√(2µ'/µ) = √(2*2πr²ρ/9πr²ρ)
→ν₀/ν₀' = √(4/9) =2/3
→ν₀ : ν₀' = 2 : 3
48. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on the left into its fundamental vibrations and that on the right into its first overtone? Take g = 10 m/s².
The figure for Q - 48
ANSWER: First we need to find the tensions in the wires for the given conditions. Let the length of each wire = L and the tensions in the left and the right wires = F and F' respectively.
For the left wire which vibrates in fundamental frequency say ν,
ν =(1/2L)*√(F/µ)
For the right wire which vibrates in 1st overtone,
frequency ν' = (2/2L)*√(F'/µ)
Since the same tuning fork is used to vibrate both of the wires,
ν = ν'
→(1/2L)*√(F/µ) = (2/2L)*√(F'/µ)
→ √F = 2√F'
→ F = 4F'
The weight of the rod 1.2 kg can be assumed to act at its CoM i.e. in the middle. Let the weight 4.8 kg be placed at x cm from the left. Consider the free body diagram of the rod as below:-
FBD of the rod
From the diagram, for the equilibrium of the rod
F+F' = W+W'
→4F' + F' = 1.2g + 4.8g
→5F' = 6g
→F' = 1.2g, and F =4*1.2g =4.8g
Since the rod is also in rotational equilibrium, the sum of torques of all the forces about a point will also be zero. Let us take the torques about the left end A.
4.8g*x + 1.2g*20 -F'*40 = 0
→4.8g*x +24g -1.2g*40 = 0
→4.8x = 48 -24
→x = 24/4.8 = 5 cm
So the mass of 4.8 kg should be kept 5 cm from the left end.
49. Figure (15-E10) shows an aluminum wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm² and that of the aluminum wire is 3.0 mm ². What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminum is 2.6 g/cm³ and that of the steel is 7.8 g/cm³.
The figure for Q - 49
ANSWER: Length of the steel wire L = 80 cm, length of the aluminum wire L' = 60 cm. Area of steel wire = 1 mm² and density =7.8 g/cm³.
The linear mass density of the steel wire
µ = 1x10⁻⁶*7.8*10⁶/1000 kg/m =7.8x10⁻³ kg/m
The area of the aluminum wire = 3 mm² and density = 2.6 g/cm³.
The linear mass density of the aluminum wire
µ' = (3x10⁻⁶)*2.6*10⁶/1000 kg/m =7.8x10⁻³ kg/m
So µ = µ' and the tension in both wire F = 40 N
Hence the wave speed = √(F/µ) =√(40/7.8x10⁻³) = 71.61 m/s
(In both of the wires)
For the minimum frequency of the tuning fork, the wavelength of the waves on the string should be maximum. For the maximum wavelength, the number of loops of the standing waves should be minimum. With the condition that the joint of the wire is a node, the maximum length of a loop will be the HCF of 80 cm and 60 cm, which is 20 cm. So there will be three number of loops in the aluminum wire and four number of loops in the steel wire. If the wavelength is 𝜆, the length of the loop = 𝜆/2.
→𝜆/2 = 20 cm
→𝜆 = 40 cm = 0.40 m
If ν is the corresponding frequency then
ν𝜆 = V
→ν = V/𝜆 = 71.61/0.40 ≈180 Hz.
50. A string of length L fixed at both ends vibrate in the fundamental mode at a frequency ν and a maximum amplitude A.
(a) Find the wavelength and the wave number k.
(b) Take the origin at one end of the string and the X-axis along the string. Take Y-axis the direction of displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-direction. Write the equation describing the standing wave.
ANSWER: (a) Here ν = V/2L, →V =2νL
Since V =νλ, so
νλ = 2νL
→λ =2L
The wave number k = 2πν/V =2πν/2νL = π/L
(b) The equation describing a standing wave in the given condition is
y = A sin kx cos ⍵t, where A = maximum amplitude, i.e. Hence
y = A sin kx cos ⍵t
→y = A sin[(π/L)x] cos[(2πν)t]
41. A wire fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz.
(a) What could be the maximum value of the fundamental frequency?
(b) If transverse waves can travel on this string at speed of 40 m/s, what is its length?
ANSWER: (a) These resonant frequencies are some nth harmonic of the fundamental. So 240 Hz and 320 Hz are some multiples of the same fundamental frequency. The maximum frequency of the fundamental will be the HCF of these two frequencies.
HCF of 240 and 320 is 80. Hence the maximum value of fundamental frequency is 80 Hz.
(b) The speed of the wave V = 40 m/s. For the fundamental frequency ν = V/2L.
→L = V/2ν =40/(2*80) m =0.25 m = 25 cm.
42. A string fixed at both ends vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency this separation is reduced to 1.6 cm. Find the length of the string.
ANSWER: Let for the nth resonant mode the separation between consecutive node is 2.0 cm. Hence the length of the string = 2n cm.
For the next higher i.e. (n+1)th resonant mode the separation is 1.6 cm. From this the length of the string = 1.6*(n+1) cm. These two expressions for the length of the string will be equal.
→2n = 1.6(n+1) =1.6n + 1.6
→0.4n = 1.6
→n =1.6/0.4 = 4
Hence the length of the string = 2n cm = 2*4 cm = 8.0 cm
43. A 660 Hz tuning fork sets up a vibration in a string clamped at both ends. The wave speed for a transverse wave on the string is 220 m/s and the string vibrates in three loops.
(a) Find the length of the string.
(b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.
ANSWER: (a) Here the wave speed V = 220 m/s. Resonant frequency ν = 660 Hz. Since the string vibrates in three loops, it is vibrating in third harmonic. i.e. n = 3. Length of the string =?
ν = nV/2L
→660 = 3*220/2L
→L = 660/(2*660) =1/2 = 0.50 m =50 cm.
(b) The equation of a wave on a string fixed at both ends is given by
y =2A sin kx cos ⍵t
Where A is the amplitude of a wave on the wire. Here,
⍵ = 2πν =2π*660 s⁻¹ =1320π s⁻¹
k =2π/λ =2πν/V, Since λ = V/ν
→k = 2π*660/220 m⁻¹ =6π m⁻¹ =0.06π cm⁻¹
The standing wave is a result of interference of two waves in opposite direction having the same amplitude A. When the standing wave is formed on the string the maximum amplitude is the sum of the amplitudes of both the waves. The sum of amplitudes of both the waves = 2A (Twice the amplitude) = Maximum amplitude =0.50 cm
Now the wave equation is
y = (0.5 cm) sin [(0.06π cm⁻¹)x]*cos[(1320π s⁻¹)t]
44. A particular guitar wire is 30 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes?
ANSWER: Here ν = 196 Hz. Since no finger is placed its vibrating length L = 30 cm. The frequency is inversely proportional to the length of the string. Hence if other things are constant,
ν/ν' = L'/L
→L' = νL/ν'
For ν' = 220 Hz
The distance from the end where the finger be placed for this frequency = 196*30/220 cm = 26.7 cm
For ν' = 247 Hz
The distance from the end where the finger be placed for this frequency = 196*30/247 cm = 23.8 cm
For ν' = 262 Hz
The distance from the end where the finger be placed for this frequency = 196*30/262 cm = 22.4 cm
For ν' = 294 Hz
The distance from the end where the finger be placed for this frequency = 196*30/294 cm = 20.0 cm
45. A steel wire fixed at both ends has a fundamental frequency of 200 Hz. A person can hear a sound of the maximum frequency of 14 kHz. What is the highest harmonic that can be played on this string which is audible to the person?
ANSWER: Here ν₀ = 200 Hz
→ V/2L = 200, {L = length of the string, V = wave speed}
Let the highest harmonic that can be played on this string and audible o the person = n. The frequency of this harmonic = νₙ
νₙ ⩽14 kHz
→nV/2L ⩽ 14000 Hz
→n*(V/2L) ⩽ 14000 Hz
→n*220 ⩽ 14000 Hz
→n ⩽ 14000/200
→n ⩽ 70
So the highest harmonic audible to the person = 70.
46. Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string.
(b) Which harmonics of the fundamental are the given frequencies?
(c) Which overtones are these frequencies?
(d) If the length of the string is 80 cm, what would be the speed of a transverse wave on the string?
ANSWER: (a) The highest possible fundamental frequency of vibration of the string will be the HCF of these three frequencies.
90 = 2*3*3*5
150 = 2*3*5*5
210 =2*3*5*7
HCF of 90, 150 and 210 Hz is =2*3*5 =30 Hz
(b) Since the higher harmonics are multiples of the fundamental harmonic,
90 Hz is the 90/30 = 3rd harmonic
150 Hz is the 150/30 =5th harmonic
210 Hz is the 210/30 =7th harmonic
(c) Since the nth harmonic is the (n-1)th overtone
3rd harmonic = 2nd overtone
5th harmonic = 4th overtone
7th harmonic = 6th overtone
(d) Given L = 80 cm =0.80 m, V = ?
For the fundamental frequency ν₀ = 30 Hz,
→V/2L = 30
→V = 30*2*0.80 = 48 m/s
47. Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2:1, the radii are in the ratio 3:1 and the densities are in the ratio 1:2. Find the ratio of their fundamental frequencies.
ANSWER: Let the tension in the wires = 2F and F and the radius = 3r and r. If the densities are ρ and 2ρ,
The linear mass density of first wire µ = π(3r)²*1*ρ =9πρr²
The linear mass density of the second wire µ' = πr²*2ρ =2πr²ρ
The fundamental frequency of the first wire ν₀ =V/2L =(1/2L)*√(2F/µ)
The fundamental frequency of the second wire ν₀' = (1/2L)*√(F/µ')
Hence the ratio of the fundamental frequencies
ν₀/ν₀' = √(2/µ)/√(1/µ') =√(2µ'/µ) = √(2*2πr²ρ/9πr²ρ)
→ν₀/ν₀' = √(4/9) =2/3
→ν₀ : ν₀' = 2 : 3
48. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on the left into its fundamental vibrations and that on the right into its first overtone? Take g = 10 m/s².
ANSWER: First we need to find the tensions in the wires for the given conditions. Let the length of each wire = L and the tensions in the left and the right wires = F and F' respectively.
For the left wire which vibrates in fundamental frequency say ν,
ν =(1/2L)*√(F/µ)
For the right wire which vibrates in 1st overtone,
frequency ν' = (2/2L)*√(F'/µ)
Since the same tuning fork is used to vibrate both of the wires,
ν = ν'
→(1/2L)*√(F/µ) = (2/2L)*√(F'/µ)
→ √F = 2√F'
→ F = 4F'
The weight of the rod 1.2 kg can be assumed to act at its CoM i.e. in the middle. Let the weight 4.8 kg be placed at x cm from the left. Consider the free body diagram of the rod as below:-
From the diagram, for the equilibrium of the rod
F+F' = W+W'
→4F' + F' = 1.2g + 4.8g
→5F' = 6g
→F' = 1.2g, and F =4*1.2g =4.8g
Since the rod is also in rotational equilibrium, the sum of torques of all the forces about a point will also be zero. Let us take the torques about the left end A.
4.8g*x + 1.2g*20 -F'*40 = 0
→4.8g*x +24g -1.2g*40 = 0
→4.8x = 48 -24
→x = 24/4.8 = 5 cm
So the mass of 4.8 kg should be kept 5 cm from the left end.
49. Figure (15-E10) shows an aluminum wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm² and that of the aluminum wire is 3.0 mm ². What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminum is 2.6 g/cm³ and that of the steel is 7.8 g/cm³.
ANSWER: Length of the steel wire L = 80 cm, length of the aluminum wire L' = 60 cm. Area of steel wire = 1 mm² and density =7.8 g/cm³.
The linear mass density of the steel wire
µ = 1x10⁻⁶*7.8*10⁶/1000 kg/m =7.8x10⁻³ kg/m
The area of the aluminum wire = 3 mm² and density = 2.6 g/cm³.
The linear mass density of the aluminum wire
µ' = (3x10⁻⁶)*2.6*10⁶/1000 kg/m =7.8x10⁻³ kg/m
So µ = µ' and the tension in both wire F = 40 N
Hence the wave speed = √(F/µ) =√(40/7.8x10⁻³) = 71.61 m/s
(In both of the wires)
For the minimum frequency of the tuning fork, the wavelength of the waves on the string should be maximum. For the maximum wavelength, the number of loops of the standing waves should be minimum. With the condition that the joint of the wire is a node, the maximum length of a loop will be the HCF of 80 cm and 60 cm, which is 20 cm. So there will be three number of loops in the aluminum wire and four number of loops in the steel wire. If the wavelength is 𝜆, the length of the loop = 𝜆/2.
→𝜆/2 = 20 cm
→𝜆 = 40 cm = 0.40 m
If ν is the corresponding frequency then
ν𝜆 = V
→ν = V/𝜆 = 71.61/0.40 ≈180 Hz.
50. A string of length L fixed at both ends vibrate in the fundamental mode at a frequency ν and a maximum amplitude A.
(a) Find the wavelength and the wave number k.
(b) Take the origin at one end of the string and the X-axis along the string. Take Y-axis the direction of displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-direction. Write the equation describing the standing wave.
ANSWER: (a) Here ν = V/2L, →V =2νL
Since V =νλ, so
νλ = 2νL
→λ =2L
The wave number k = 2πν/V =2πν/2νL = π/L
(b) The equation describing a standing wave in the given condition is
y = A sin kx cos ⍵t, where A = maximum amplitude, i.e. Hence
y = A sin kx cos ⍵t
→y = A sin[(π/L)x] cos[(2πν)t]
(a) What could be the maximum value of the fundamental frequency?
(b) If transverse waves can travel on this string at speed of 40 m/s, what is its length?
ANSWER: (a) These resonant frequencies are some nth harmonic of the fundamental. So 240 Hz and 320 Hz are some multiples of the same fundamental frequency. The maximum frequency of the fundamental will be the HCF of these two frequencies.
HCF of 240 and 320 is 80. Hence the maximum value of fundamental frequency is 80 Hz.
(b) The speed of the wave V = 40 m/s. For the fundamental frequency ν = V/2L.
→L = V/2ν =40/(2*80) m =0.25 m = 25 cm.
42. A string fixed at both ends vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency this separation is reduced to 1.6 cm. Find the length of the string.
ANSWER: Let for the nth resonant mode the separation between consecutive node is 2.0 cm. Hence the length of the string = 2n cm.
For the next higher i.e. (n+1)th resonant mode the separation is 1.6 cm. From this the length of the string = 1.6*(n+1) cm. These two expressions for the length of the string will be equal.
→2n = 1.6(n+1) =1.6n + 1.6
→0.4n = 1.6
→n =1.6/0.4 = 4
Hence the length of the string = 2n cm = 2*4 cm = 8.0 cm
43. A 660 Hz tuning fork sets up a vibration in a string clamped at both ends. The wave speed for a transverse wave on the string is 220 m/s and the string vibrates in three loops.
(a) Find the length of the string.
(b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.
ANSWER: (a) Here the wave speed V = 220 m/s. Resonant frequency ν = 660 Hz. Since the string vibrates in three loops, it is vibrating in third harmonic. i.e. n = 3. Length of the string =?
ν = nV/2L
→660 = 3*220/2L
→L = 660/(2*660) =1/2 = 0.50 m =50 cm.
(b) The equation of a wave on a string fixed at both ends is given by
y =2A sin kx cos ⍵t
Where A is the amplitude of a wave on the wire. Here,
⍵ = 2πν =2π*660 s⁻¹ =1320π s⁻¹
k =2π/λ =2πν/V, Since λ = V/ν
→k = 2π*660/220 m⁻¹ =6π m⁻¹ =0.06π cm⁻¹
The standing wave is a result of interference of two waves in opposite direction having the same amplitude A. When the standing wave is formed on the string the maximum amplitude is the sum of the amplitudes of both the waves. The sum of amplitudes of both the waves = 2A (Twice the amplitude) = Maximum amplitude =0.50 cm
Now the wave equation is
y = (0.5 cm) sin [(0.06π cm⁻¹)x]*cos[(1320π s⁻¹)t]
44. A particular guitar wire is 30 cm long and vibrates at a frequency of 196 Hz when no finger is placed on it. The next higher notes on the scale are 220 Hz, 247 Hz, 262 Hz and 294 Hz. How far from the end of the string must the finger be placed to play these notes?
ANSWER: Here ν = 196 Hz. Since no finger is placed its vibrating length L = 30 cm. The frequency is inversely proportional to the length of the string. Hence if other things are constant,
ν/ν' = L'/L
→L' = νL/ν'
For ν' = 220 Hz
The distance from the end where the finger be placed for this frequency = 196*30/220 cm = 26.7 cm
For ν' = 247 Hz
The distance from the end where the finger be placed for this frequency = 196*30/247 cm = 23.8 cm
For ν' = 262 Hz
The distance from the end where the finger be placed for this frequency = 196*30/262 cm = 22.4 cm
For ν' = 294 Hz
The distance from the end where the finger be placed for this frequency = 196*30/294 cm = 20.0 cm
ANSWER: Here ν₀ = 200 Hz
→ V/2L = 200, {L = length of the string, V = wave speed}
Let the highest harmonic that can be played on this string and audible o the person = n. The frequency of this harmonic = νₙ
νₙ ⩽14 kHz
→nV/2L ⩽ 14000 Hz
→n*(V/2L) ⩽ 14000 Hz
→n*220 ⩽ 14000 Hz
→n ⩽ 14000/200
→n ⩽ 70
So the highest harmonic audible to the person = 70.
46. Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string.
(b) Which harmonics of the fundamental are the given frequencies?
(c) Which overtones are these frequencies?
(d) If the length of the string is 80 cm, what would be the speed of a transverse wave on the string?
ANSWER: (a) The highest possible fundamental frequency of vibration of the string will be the HCF of these three frequencies.
90 = 2*3*3*5
150 = 2*3*5*5
210 =2*3*5*7
HCF of 90, 150 and 210 Hz is =2*3*5 =30 Hz
(b) Since the higher harmonics are multiples of the fundamental harmonic,
90 Hz is the 90/30 = 3rd harmonic
150 Hz is the 150/30 =5th harmonic
210 Hz is the 210/30 =7th harmonic
(c) Since the nth harmonic is the (n-1)th overtone
3rd harmonic = 2nd overtone
5th harmonic = 4th overtone
7th harmonic = 6th overtone
(d) Given L = 80 cm =0.80 m, V = ?
For the fundamental frequency ν₀ = 30 Hz,
→V/2L = 30
→V = 30*2*0.80 = 48 m/s
47. Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2:1, the radii are in the ratio 3:1 and the densities are in the ratio 1:2. Find the ratio of their fundamental frequencies.
ANSWER: Let the tension in the wires = 2F and F and the radius = 3r and r. If the densities are ρ and 2ρ,
The linear mass density of first wire µ = π(3r)²*1*ρ =9πρr²
The linear mass density of the second wire µ' = πr²*2ρ =2πr²ρ
The fundamental frequency of the first wire ν₀ =V/2L =(1/2L)*√(2F/µ)
The fundamental frequency of the second wire ν₀' = (1/2L)*√(F/µ')
Hence the ratio of the fundamental frequencies
ν₀/ν₀' = √(2/µ)/√(1/µ') =√(2µ'/µ) = √(2*2πr²ρ/9πr²ρ)
→ν₀/ν₀' = √(4/9) =2/3
→ν₀ : ν₀' = 2 : 3
48. A uniform horizontal rod of length 40 cm and mass 1.2 kg is supported by two identical wires as shown in figure (15-E9). Where should a mass of 4.8 kg be placed on the rod so that the same tuning fork may excite the wire on the left into its fundamental vibrations and that on the right into its first overtone? Take g = 10 m/s².
|
| The figure for Q - 48 |
ANSWER: First we need to find the tensions in the wires for the given conditions. Let the length of each wire = L and the tensions in the left and the right wires = F and F' respectively.
For the left wire which vibrates in fundamental frequency say ν,
ν =(1/2L)*√(F/µ)
For the right wire which vibrates in 1st overtone,
frequency ν' = (2/2L)*√(F'/µ)
Since the same tuning fork is used to vibrate both of the wires,
ν = ν'
→(1/2L)*√(F/µ) = (2/2L)*√(F'/µ)
→ √F = 2√F'
→ F = 4F'
The weight of the rod 1.2 kg can be assumed to act at its CoM i.e. in the middle. Let the weight 4.8 kg be placed at x cm from the left. Consider the free body diagram of the rod as below:-
|
| FBD of the rod |
From the diagram, for the equilibrium of the rod
F+F' = W+W'
→4F' + F' = 1.2g + 4.8g
→5F' = 6g
→F' = 1.2g, and F =4*1.2g =4.8g
Since the rod is also in rotational equilibrium, the sum of torques of all the forces about a point will also be zero. Let us take the torques about the left end A.
4.8g*x + 1.2g*20 -F'*40 = 0
→4.8g*x +24g -1.2g*40 = 0
→4.8x = 48 -24
→x = 24/4.8 = 5 cm
So the mass of 4.8 kg should be kept 5 cm from the left end.
49. Figure (15-E10) shows an aluminum wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is 1.0 mm² and that of the aluminum wire is 3.0 mm ². What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node? The density of aluminum is 2.6 g/cm³ and that of the steel is 7.8 g/cm³.
 |
| The figure for Q - 49 |
ANSWER: Length of the steel wire L = 80 cm, length of the aluminum wire L' = 60 cm. Area of steel wire = 1 mm² and density =7.8 g/cm³.
The linear mass density of the steel wire
µ = 1x10⁻⁶*7.8*10⁶/1000 kg/m =7.8x10⁻³ kg/m
The area of the aluminum wire = 3 mm² and density = 2.6 g/cm³.
The linear mass density of the aluminum wire
µ' = (3x10⁻⁶)*2.6*10⁶/1000 kg/m =7.8x10⁻³ kg/m
So µ = µ' and the tension in both wire F = 40 N
Hence the wave speed = √(F/µ) =√(40/7.8x10⁻³) = 71.61 m/s
(In both of the wires)
For the minimum frequency of the tuning fork, the wavelength of the waves on the string should be maximum. For the maximum wavelength, the number of loops of the standing waves should be minimum. With the condition that the joint of the wire is a node, the maximum length of a loop will be the HCF of 80 cm and 60 cm, which is 20 cm. So there will be three number of loops in the aluminum wire and four number of loops in the steel wire. If the wavelength is 𝜆, the length of the loop = 𝜆/2.
→𝜆/2 = 20 cm
→𝜆 = 40 cm = 0.40 m
If ν is the corresponding frequency then
ν𝜆 = V
→ν = V/𝜆 = 71.61/0.40 ≈180 Hz.
50. A string of length L fixed at both ends vibrate in the fundamental mode at a frequency ν and a maximum amplitude A.
(a) Find the wavelength and the wave number k.
(b) Take the origin at one end of the string and the X-axis along the string. Take Y-axis the direction of displacement. Take t = 0 at the instant when the middle point of the string passes through its mean position and is going towards the positive y-direction. Write the equation describing the standing wave.
ANSWER: (a) Here ν = V/2L, →V =2νL
Since V =νλ, so
νλ = 2νL
→λ =2L
The wave number k = 2πν/V =2πν/2νL = π/L
(b) The equation describing a standing wave in the given condition is
y = A sin kx cos ⍵t, where A = maximum amplitude, i.e. Hence
y = A sin kx cos ⍵t
→y = A sin[(π/L)x] cos[(2πν)t]
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 11 - Gravitation
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CHAPTER- 10 - Rotational Mechanics
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CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 11 - Gravitation
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
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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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