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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
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
(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.
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?
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
(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.
at x = 0, x = 10 cm, x = 20 cm and x = 30 cm.
 |
| The diagram for Q - 52 |
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
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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
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
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
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
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
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 → 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)
Click here for → Exercises (21-30)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
Click here for → Exercises (31-42)
Click here for → Exercise(43-54)
CHAPTER- 7 - Circular Motion
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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 → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
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
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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".
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)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
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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