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WAVE MOTION AND WAVES ON A STRING
EXERCISES:- Q-11 to Q-20
11. A particle on a stretched string supporting a traveling wave takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.
ANSWER: Since the distance between the two consecutive particles which are at their mean position = 2.0 cm, it means it is equal to half the wavelength.
∴ 𝜆/2 = 2.0 cm
→𝜆 = 4.0 cm = Wavelength
The time taken by a particle to reach the extreme from the mean position is one-fourth of the time period T. Hence,
T/4 = 5.0 ms
→T = 20 ms = 20x10⁻³ s
∴ The frequency ν = 1/T = 1/20x10⁻³ Hz
→ν = 1000/20 Hz =50 Hz
The wave speed v = ν𝜆 =50*4 cm/s =200/100 m/s = 2.0 m/s
12. Figure (15-E2) shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm/s. Find (a) the amplitude, (b) the wavelength, (c) the wave number and the frequency of the wave.
Figure for Q-12
ANSWER: (a) The amplitude of the wave is the maximum transverse displacement. From the given figure the amplitude is = 1.0 mm.
(b) The wavelength is the distance between two consecutive particles along the propagation of the wave which are in the same phase. From the given figure, wavelength = 4 cm.
(c) Wave speed v = ν𝜆 {Where 𝜆 is the wavelength and ν i the frequency.}
→ν = v/𝜆 =(20 cm/s)/(4 cm) =5 Hz
The wave number k = 2π/𝜆 = 2*3.14/4 ≈1.6 cm⁻¹.
13. A wave traveling on a string at a speed of 10 m/s causes each particle of the string to oscillate with a time period of 20 ms.
(a) What is the wavelength of the wave?
(b) If the displacement of a particle is 1.5 mm at a certain instant, what will be the displacement of a particle 10 cm away from it at the same instant?
ANSWER: (a) Given v = 10 m/s.
The time period of the oscillation, T = 20 ms =20/1000 s =0.02 s
But v = 𝜆/T Where 𝜆 is the wavelength.
→𝜆 = vT =10*0.02 m =0.2 m = 20 cm
(b) Given y = 1.5 mm. Since the displacement required is at 𝜆/2 distance, the particle in consideration is at T/2 time away. Let the displacement of the particle in question = y'.
y'/y = sin⍵(t+T/2)/sin⍵t = sin(⍵t+ωT/2)/sin⍵t
But T=2π/⍵ →⍵T/2 =π
Now y'/y = sin(⍵t+π)/sin⍵t = -sin⍵t/sin⍵t = -1
→y' = -y = -1.5 mm
14. A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it?
ANSWER: Length = 64 cm =0.64 m
Weight = 5 g = 0.005 kg
Linear mass density, µ = 0.005/0.64 kg/m = 0.0078 kg/m
Tension in the string F = 8 N
Hence the velocity of the wave on the string, v = √(F/µ)
→v = √(8/0.0078) m/s = √1025.64 m/s = 32 m/s.
15. A string of length 20 cm and linear mass density of 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an end as shown in figure (15-E3), which travels towards the other end.
(a) When will the string have the shape shown in the figure again?
(b) Sketch the shape of the string at a time half of that found in part (a).
The figure for Q-15
ANSWER: (a) Given µ = 0.40 g/cm =0.40*100/1000 kg/m = 0.04 kg/m
and F = 16 N. Hence the wave speed v =√(F/µ)
→v = √(16/0.04) =√400 =20 m/s. The pulse will reflect back from the fixed end at right but its shape will be inverted. When the pulse will reflect back again from the left end, the pulse will again become similar to the initial shape and the string will get the shape shown in the figure. Thus the pulse will have to travel twice the length of the string = 2*20 cm = 40 cm = 0.40 m. With a wave speed v = 20 m/s, the time taken to travel this distance
= (0.40 m)/(20 m/s) =0.02 s
(b) In half of this time, the pulse will travel half the distance traveled in part (a) i.e. =0.20 m which is the full length of the string. At this time the left end of the pulse will be at the right fixed end and the shape of the pulse will be inverted. The sketch of the shape of the string will be as follows:-
Diagram for Q-15
16. A string of mass density of 0.50 g/cm and a total length of 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure 15-E4). The ring can move on a vertical rod. A wave pulse is on the string which moves towards the ring at a speed of 20 cm/s. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring.
(a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape.
(b) The shape of the string changes periodically with time. Find this time period
(c) What is the tension in the string?
The figure for Q-16
ANSWER: (a) The speed of the pulse, v = 20 cm/s = 0.20 m/s.
The pulse will reflect back at the ring-end regaining its shape and travel again 20 cm to give the string its initial shape. Total distance traveled = 2*20 cm = 40 cm = 0.40 m.
Time taken = (0.40 m)/(0.20 m/s) = 2 s.
(b) The shape of the string changes when the pulse reflects at the fixed end. The pulse needs a full to and fro movement on the string from the left end to change its shape. Thus the distance traveled = 2*30 cm = 60 cm =0.60 m. Required time period = time taken to travel 0.60 m = (0.60 m)/(0.20 m/s) = 3 s.
(c) Given µ = 0.50 g/cm = 0.50*100/1000 kg/m =0.05 kg/m
v = 0.20 m/s. Hence v = √(F/µ) {F = Tension in the string}
→0.20 = √(F/0.05)
→F = 0.20² * 0.05 N = 0.04*0.05 N
→F = 0.0020 N = 2 x 10⁻³ N.
17. Two wires of different densities but same area of the cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.
ANSWER: Let the density of the first wire ρ and linear mass density =µ. Similarly the density of the second wire ρ' and linear mass density =µ'. Since the cross-section area is the same,
µ/µ' = ρ/ρ'
If the velocity of the transverse wave in the second wire = v, then
the velocity of the wave in the first wire = 2v. The tension T is the same in both the wires. Hence,
2v =√(T/µ) and v = √(T/µ')
Thus 2*√(T/µ') = √(T/µ)
→µ'/4 = µ
→µ/µ' = 1/4
→ρ/ρ' = 1/4 = 0.25
18. A transverse wave described by
y = (0.02 m) sin[(1.0 m-1)x +(30 s-1)t]
propagates on a stretched string having a linear mass density of 1.2x10-4 kg/m. Find the tension in the string.
ANSWER: Given µ = 1.2 x 10⁻⁴ kg/m.
And y = (0.02 m) sin[(1.0 m⁻¹)x + (30 s⁻¹)t]
Comparing it with the wave equation
y = A sin[⍵t - ⍵x/v]
We get, ⍵ = 30 s⁻¹ and ⍵/v = -1.0 m⁻¹
So the wave speed v = (30 s⁻¹)/(-1.0 m⁻¹) =-30 m/s
But v =√(F/µ) {where F = tension in the string}
→ -30 = √(F/µ)
→900 = F/(1.2x10⁻⁴)
→F = 900*(1.2x10⁻⁴) =0.108 N
19. A traveling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of the vibration is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10 kg/m and is kept under a tension of 90 N.
(a) Find the speed and the wavelength of the wave.
(b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation.
(c) Find the velocity and the acceleration of the particle at x = 50 cm at time t = 10 ms.
ANSWER: Given µ = 0.10 kg/m and A = 1 cm = 0.01 m, F = 90 N.
Since the displacement becomes zero twice during a cycle, the frequency of the wave, ν = 200/2 =100 Hz.
(a) Wave speed v =√(F/µ) =√(90/0.10) =√900 = 30 m/s
Also v =ν𝜆
→30 =100*𝜆
→𝜆 = 30/100 m =30 cm.
(b) General wave equation
y = A sin[⍵t-kx+ⲫ], at t = 0, x = 0, y = A
→A = A sinⲫ
→sinⲫ = 1
→ⲫ = π/2
⍵ = 2πν =2π*100 =200π s⁻¹
k =⍵/v = 200π/30 = 20π/3 m⁻¹ =20π/300 cm⁻¹ =2π/30 cm⁻¹
Hence the wave equation is
y = (1.0 cm) sin [(200π s⁻¹)t - (2π/30 cm⁻¹)x+π/2]
→y = (1.0 cm) cos [(200π s⁻¹)t - (2π/30 cm⁻¹)x]
→y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
(c) From the equation
y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
Partially differentiating with respet to time we get,
v = ∂y/∂t = -(1.0 cm) (2π/0.01) sin 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
At x = 50 cm, t = 10 ms =0.01 s
v = -200π sin2π[1-(50/30)] =-200π sin(-4π/3) = 200π sin(π+π/3)
→v = -200π sin(π/3) =-628√3/2 cm/s = -543 cm/s
→v = -5.43 m/s
Acceleration a =∂v/∂t
=-(1.0 cm)(2π/0.01)² cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
At x = 50 cm, t = 10 ms =0.01 s
a = -(200π)² cos 2π[1-50/30] =-(200π)² cos[-4π/3]
→a =-(200π)²cos(π+π/3) = (200π)²*cosπ/3
→a = (200π)²*½ cm/s² =20000π² cm/s² =0.20*π² km/s²
→a = 1.97 km/s² ≈ 2 km/s²
20. A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160 N/m and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring?
ANSWER: Elongation of the spring, x = 1 cm = 0.01 m
Spring constant k = 160 N/m.
Force on the spring F =kx =160*0.01 N =1.6 N
µ = (10 g)/(40 cm) =0.01/0.4 kg/m = 0.025 kg/m
Wave speed v = √(F/µ) =√(1.6/0.025) =√(64) = 8 m/s
Length of the string L = 40 cm = 0.40 m
Hence time taken by the pulse to reach the string =L/v
=0.40/8 s = 0.05 s
11. A particle on a stretched string supporting a traveling wave takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.
ANSWER: Since the distance between the two consecutive particles which are at their mean position = 2.0 cm, it means it is equal to half the wavelength.
∴ 𝜆/2 = 2.0 cm
→𝜆 = 4.0 cm = Wavelength
The time taken by a particle to reach the extreme from the mean position is one-fourth of the time period T. Hence,
T/4 = 5.0 ms
→T = 20 ms = 20x10⁻³ s
∴ The frequency ν = 1/T = 1/20x10⁻³ Hz
→ν = 1000/20 Hz =50 Hz
The wave speed v = ν𝜆 =50*4 cm/s =200/100 m/s = 2.0 m/s
12. Figure (15-E2) shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm/s. Find (a) the amplitude, (b) the wavelength, (c) the wave number and the frequency of the wave.
ANSWER: (a) The amplitude of the wave is the maximum transverse displacement. From the given figure the amplitude is = 1.0 mm.
(b) The wavelength is the distance between two consecutive particles along the propagation of the wave which are in the same phase. From the given figure, wavelength = 4 cm.
(c) Wave speed v = ν𝜆 {Where 𝜆 is the wavelength and ν i the frequency.}
→ν = v/𝜆 =(20 cm/s)/(4 cm) =5 Hz
The wave number k = 2π/𝜆 = 2*3.14/4 ≈1.6 cm⁻¹.
ANSWER: Since the distance between the two consecutive particles which are at their mean position = 2.0 cm, it means it is equal to half the wavelength.
∴ 𝜆/2 = 2.0 cm
→𝜆 = 4.0 cm = Wavelength
The time taken by a particle to reach the extreme from the mean position is one-fourth of the time period T. Hence,
T/4 = 5.0 ms
→T = 20 ms = 20x10⁻³ s
∴ The frequency ν = 1/T = 1/20x10⁻³ Hz
→ν = 1000/20 Hz =50 Hz
The wave speed v = ν𝜆 =50*4 cm/s =200/100 m/s = 2.0 m/s
12. Figure (15-E2) shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm/s. Find (a) the amplitude, (b) the wavelength, (c) the wave number and the frequency of the wave.
 |
| Figure for Q-12 |
ANSWER: (a) The amplitude of the wave is the maximum transverse displacement. From the given figure the amplitude is = 1.0 mm.
(b) The wavelength is the distance between two consecutive particles along the propagation of the wave which are in the same phase. From the given figure, wavelength = 4 cm.
(c) Wave speed v = ν𝜆 {Where 𝜆 is the wavelength and ν i the frequency.}
→ν = v/𝜆 =(20 cm/s)/(4 cm) =5 Hz
The wave number k = 2π/𝜆 = 2*3.14/4 ≈1.6 cm⁻¹.
13. A wave traveling on a string at a speed of 10 m/s causes each particle of the string to oscillate with a time period of 20 ms.
(a) What is the wavelength of the wave?
(b) If the displacement of a particle is 1.5 mm at a certain instant, what will be the displacement of a particle 10 cm away from it at the same instant?
ANSWER: (a) Given v = 10 m/s.
The time period of the oscillation, T = 20 ms =20/1000 s =0.02 s
But v = 𝜆/T Where 𝜆 is the wavelength.
→𝜆 = vT =10*0.02 m =0.2 m = 20 cm
(b) Given y = 1.5 mm. Since the displacement required is at 𝜆/2 distance, the particle in consideration is at T/2 time away. Let the displacement of the particle in question = y'.
y'/y = sin⍵(t+T/2)/sin⍵t = sin(⍵t+ωT/2)/sin⍵t
But T=2π/⍵ →⍵T/2 =π
Now y'/y = sin(⍵t+π)/sin⍵t = -sin⍵t/sin⍵t = -1
→y' = -y = -1.5 mm
14. A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it?
ANSWER: Length = 64 cm =0.64 m
Weight = 5 g = 0.005 kg
Linear mass density, µ = 0.005/0.64 kg/m = 0.0078 kg/m
Tension in the string F = 8 N
Hence the velocity of the wave on the string, v = √(F/µ)
→v = √(8/0.0078) m/s = √1025.64 m/s = 32 m/s.
15. A string of length 20 cm and linear mass density of 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an end as shown in figure (15-E3), which travels towards the other end.
(a) When will the string have the shape shown in the figure again?
(b) Sketch the shape of the string at a time half of that found in part (a).
ANSWER: (a) Given µ = 0.40 g/cm =0.40*100/1000 kg/m = 0.04 kg/m
and F = 16 N. Hence the wave speed v =√(F/µ)
→v = √(16/0.04) =√400 =20 m/s. The pulse will reflect back from the fixed end at right but its shape will be inverted. When the pulse will reflect back again from the left end, the pulse will again become similar to the initial shape and the string will get the shape shown in the figure. Thus the pulse will have to travel twice the length of the string = 2*20 cm = 40 cm = 0.40 m. With a wave speed v = 20 m/s, the time taken to travel this distance
= (0.40 m)/(20 m/s) =0.02 s
(b) In half of this time, the pulse will travel half the distance traveled in part (a) i.e. =0.20 m which is the full length of the string. At this time the left end of the pulse will be at the right fixed end and the shape of the pulse will be inverted. The sketch of the shape of the string will be as follows:-
(a) What is the wavelength of the wave?
(b) If the displacement of a particle is 1.5 mm at a certain instant, what will be the displacement of a particle 10 cm away from it at the same instant?
ANSWER: (a) Given v = 10 m/s.
The time period of the oscillation, T = 20 ms =20/1000 s =0.02 s
But v = 𝜆/T Where 𝜆 is the wavelength.
→𝜆 = vT =10*0.02 m =0.2 m = 20 cm
(b) Given y = 1.5 mm. Since the displacement required is at 𝜆/2 distance, the particle in consideration is at T/2 time away. Let the displacement of the particle in question = y'.
y'/y = sin⍵(t+T/2)/sin⍵t = sin(⍵t+ωT/2)/sin⍵t
But T=2π/⍵ →⍵T/2 =π
Now y'/y = sin(⍵t+π)/sin⍵t = -sin⍵t/sin⍵t = -1
→y' = -y = -1.5 mm
14. A steel wire of length 64 cm weighs 5 g. If it is stretched by a force of 8 N, what would be the speed of a transverse wave passing on it?
ANSWER: Length = 64 cm =0.64 m
Weight = 5 g = 0.005 kg
Linear mass density, µ = 0.005/0.64 kg/m = 0.0078 kg/m
Tension in the string F = 8 N
Hence the velocity of the wave on the string, v = √(F/µ)
→v = √(8/0.0078) m/s = √1025.64 m/s = 32 m/s.
15. A string of length 20 cm and linear mass density of 0.40 g/cm is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an end as shown in figure (15-E3), which travels towards the other end.
(a) When will the string have the shape shown in the figure again?
(b) Sketch the shape of the string at a time half of that found in part (a).
 |
| The figure for Q-15 |
ANSWER: (a) Given µ = 0.40 g/cm =0.40*100/1000 kg/m = 0.04 kg/m
and F = 16 N. Hence the wave speed v =√(F/µ)
→v = √(16/0.04) =√400 =20 m/s. The pulse will reflect back from the fixed end at right but its shape will be inverted. When the pulse will reflect back again from the left end, the pulse will again become similar to the initial shape and the string will get the shape shown in the figure. Thus the pulse will have to travel twice the length of the string = 2*20 cm = 40 cm = 0.40 m. With a wave speed v = 20 m/s, the time taken to travel this distance
= (0.40 m)/(20 m/s) =0.02 s
(b) In half of this time, the pulse will travel half the distance traveled in part (a) i.e. =0.20 m which is the full length of the string. At this time the left end of the pulse will be at the right fixed end and the shape of the pulse will be inverted. The sketch of the shape of the string will be as follows:-
 |
| Diagram for Q-15 |
16. A string of mass density of 0.50 g/cm and a total length of 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end (figure 15-E4). The ring can move on a vertical rod. A wave pulse is on the string which moves towards the ring at a speed of 20 cm/s. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring.
(a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape.
(b) The shape of the string changes periodically with time. Find this time period
(c) What is the tension in the string?
ANSWER: (a) The speed of the pulse, v = 20 cm/s = 0.20 m/s.
The pulse will reflect back at the ring-end regaining its shape and travel again 20 cm to give the string its initial shape. Total distance traveled = 2*20 cm = 40 cm = 0.40 m.
Time taken = (0.40 m)/(0.20 m/s) = 2 s.
(b) The shape of the string changes when the pulse reflects at the fixed end. The pulse needs a full to and fro movement on the string from the left end to change its shape. Thus the distance traveled = 2*30 cm = 60 cm =0.60 m. Required time period = time taken to travel 0.60 m = (0.60 m)/(0.20 m/s) = 3 s.
(c) Given µ = 0.50 g/cm = 0.50*100/1000 kg/m =0.05 kg/m
v = 0.20 m/s. Hence v = √(F/µ) {F = Tension in the string}
→0.20 = √(F/0.05)
→F = 0.20² * 0.05 N = 0.04*0.05 N
→F = 0.0020 N = 2 x 10⁻³ N.
17. Two wires of different densities but same area of the cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.
ANSWER: Let the density of the first wire ρ and linear mass density =µ. Similarly the density of the second wire ρ' and linear mass density =µ'. Since the cross-section area is the same,
µ/µ' = ρ/ρ'
If the velocity of the transverse wave in the second wire = v, then
the velocity of the wave in the first wire = 2v. The tension T is the same in both the wires. Hence,
2v =√(T/µ) and v = √(T/µ')
Thus 2*√(T/µ') = √(T/µ)
→µ'/4 = µ
→µ/µ' = 1/4
→ρ/ρ' = 1/4 = 0.25
(a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape.
(b) The shape of the string changes periodically with time. Find this time period
(c) What is the tension in the string?
 |
| The figure for Q-16 |
ANSWER: (a) The speed of the pulse, v = 20 cm/s = 0.20 m/s.
The pulse will reflect back at the ring-end regaining its shape and travel again 20 cm to give the string its initial shape. Total distance traveled = 2*20 cm = 40 cm = 0.40 m.
Time taken = (0.40 m)/(0.20 m/s) = 2 s.
(b) The shape of the string changes when the pulse reflects at the fixed end. The pulse needs a full to and fro movement on the string from the left end to change its shape. Thus the distance traveled = 2*30 cm = 60 cm =0.60 m. Required time period = time taken to travel 0.60 m = (0.60 m)/(0.20 m/s) = 3 s.
(c) Given µ = 0.50 g/cm = 0.50*100/1000 kg/m =0.05 kg/m
v = 0.20 m/s. Hence v = √(F/µ) {F = Tension in the string}
→0.20 = √(F/0.05)
→F = 0.20² * 0.05 N = 0.04*0.05 N
→F = 0.0020 N = 2 x 10⁻³ N.
17. Two wires of different densities but same area of the cross-section are soldered together at one end and are stretched to a tension T. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.
ANSWER: Let the density of the first wire ρ and linear mass density =µ. Similarly the density of the second wire ρ' and linear mass density =µ'. Since the cross-section area is the same,
µ/µ' = ρ/ρ'
If the velocity of the transverse wave in the second wire = v, then
the velocity of the wave in the first wire = 2v. The tension T is the same in both the wires. Hence,
2v =√(T/µ) and v = √(T/µ')
Thus 2*√(T/µ') = √(T/µ)
→µ'/4 = µ
→µ/µ' = 1/4
→ρ/ρ' = 1/4 = 0.25
18. A transverse wave described by
y = (0.02 m) sin[(1.0 m-1)x +(30 s-1)t]
propagates on a stretched string having a linear mass density of 1.2x10-4 kg/m. Find the tension in the string.
ANSWER: Given µ = 1.2 x 10⁻⁴ kg/m.
And y = (0.02 m) sin[(1.0 m⁻¹)x + (30 s⁻¹)t]
Comparing it with the wave equation
y = A sin[⍵t - ⍵x/v]
We get, ⍵ = 30 s⁻¹ and ⍵/v = -1.0 m⁻¹
So the wave speed v = (30 s⁻¹)/(-1.0 m⁻¹) =-30 m/s
But v =√(F/µ) {where F = tension in the string}
→ -30 = √(F/µ)
→900 = F/(1.2x10⁻⁴)
→F = 900*(1.2x10⁻⁴) =0.108 N
19. A traveling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of the vibration is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10 kg/m and is kept under a tension of 90 N.
(a) Find the speed and the wavelength of the wave.
(b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation.
(c) Find the velocity and the acceleration of the particle at x = 50 cm at time t = 10 ms.
ANSWER: Given µ = 0.10 kg/m and A = 1 cm = 0.01 m, F = 90 N.
Since the displacement becomes zero twice during a cycle, the frequency of the wave, ν = 200/2 =100 Hz.
(a) Wave speed v =√(F/µ) =√(90/0.10) =√900 = 30 m/s
Also v =ν𝜆
→30 =100*𝜆
→𝜆 = 30/100 m =30 cm.
(b) General wave equation
y = A sin[⍵t-kx+ⲫ], at t = 0, x = 0, y = A
→A = A sinⲫ
→sinⲫ = 1
→ⲫ = π/2
⍵ = 2πν =2π*100 =200π s⁻¹
k =⍵/v = 200π/30 = 20π/3 m⁻¹ =20π/300 cm⁻¹ =2π/30 cm⁻¹
Hence the wave equation is
y = (1.0 cm) sin [(200π s⁻¹)t - (2π/30 cm⁻¹)x+π/2]
→y = (1.0 cm) cos [(200π s⁻¹)t - (2π/30 cm⁻¹)x]
→y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
(c) From the equation
y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
Partially differentiating with respet to time we get,
v = ∂y/∂t = -(1.0 cm) (2π/0.01) sin 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
y = (0.02 m) sin[(1.0 m-1)x +(30 s-1)t]
propagates on a stretched string having a linear mass density of 1.2x10-4 kg/m. Find the tension in the string.
ANSWER: Given µ = 1.2 x 10⁻⁴ kg/m.
And y = (0.02 m) sin[(1.0 m⁻¹)x + (30 s⁻¹)t]
Comparing it with the wave equation
y = A sin[⍵t - ⍵x/v]
We get, ⍵ = 30 s⁻¹ and ⍵/v = -1.0 m⁻¹
So the wave speed v = (30 s⁻¹)/(-1.0 m⁻¹) =-30 m/s
But v =√(F/µ) {where F = tension in the string}
→ -30 = √(F/µ)
→900 = F/(1.2x10⁻⁴)
→F = 900*(1.2x10⁻⁴) =0.108 N
19. A traveling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of the vibration is 1.0 cm and the displacement becomes zero 200 times per second. The linear mass density of the string is 0.10 kg/m and is kept under a tension of 90 N.
(a) Find the speed and the wavelength of the wave.
(b) Assume that the wave moves in the positive x-direction and at t = 0, the end x = 0 is at its positive extreme position. Write the wave equation.
(c) Find the velocity and the acceleration of the particle at x = 50 cm at time t = 10 ms.
ANSWER: Given µ = 0.10 kg/m and A = 1 cm = 0.01 m, F = 90 N.
Since the displacement becomes zero twice during a cycle, the frequency of the wave, ν = 200/2 =100 Hz.
(a) Wave speed v =√(F/µ) =√(90/0.10) =√900 = 30 m/s
Also v =ν𝜆
→30 =100*𝜆
→𝜆 = 30/100 m =30 cm.
(b) General wave equation
y = A sin[⍵t-kx+ⲫ], at t = 0, x = 0, y = A
→A = A sinⲫ
→sinⲫ = 1
→ⲫ = π/2
⍵ = 2πν =2π*100 =200π s⁻¹
k =⍵/v = 200π/30 = 20π/3 m⁻¹ =20π/300 cm⁻¹ =2π/30 cm⁻¹
Hence the wave equation is
y = (1.0 cm) sin [(200π s⁻¹)t - (2π/30 cm⁻¹)x+π/2]
→y = (1.0 cm) cos [(200π s⁻¹)t - (2π/30 cm⁻¹)x]
→y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
(c) From the equation
y =(1.0 cm) cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
Partially differentiating with respet to time we get,
v = ∂y/∂t = -(1.0 cm) (2π/0.01) sin 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
At x = 50 cm, t = 10 ms =0.01 s
v = -200π sin2π[1-(50/30)] =-200π sin(-4π/3) = 200π sin(π+π/3)
→v = -200π sin(π/3) =-628√3/2 cm/s = -543 cm/s
→v = -5.43 m/s
Acceleration a =∂v/∂t
=-(1.0 cm)(2π/0.01)² cos 2π[(t/0.01 s⁻¹) - (x/30 cm⁻¹)]
At x = 50 cm, t = 10 ms =0.01 s
a = -(200π)² cos 2π[1-50/30] =-(200π)² cos[-4π/3]
→a =-(200π)²cos(π+π/3) = (200π)²*cosπ/3
→a = (200π)²*½ cm/s² =20000π² cm/s² =0.20*π² km/s²
→a = 1.97 km/s² ≈ 2 km/s²
20. A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of 160 N/m and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring?
ANSWER: Elongation of the spring, x = 1 cm = 0.01 m
Spring constant k = 160 N/m.
Force on the spring F =kx =160*0.01 N =1.6 N
µ = (10 g)/(40 cm) =0.01/0.4 kg/m = 0.025 kg/m
Wave speed v = √(F/µ) =√(1.6/0.025) =√(64) = 8 m/s
Length of the string L = 40 cm = 0.40 m
Hence time taken by the pulse to reach the string =L/v
=0.40/8 s = 0.05 s
ANSWER: Elongation of the spring, x = 1 cm = 0.01 m
Spring constant k = 160 N/m.
Force on the spring F =kx =160*0.01 N =1.6 N
µ = (10 g)/(40 cm) =0.01/0.4 kg/m = 0.025 kg/m
Wave speed v = √(F/µ) =√(1.6/0.025) =√(64) = 8 m/s
Length of the string L = 40 cm = 0.40 m
Hence time taken by the pulse to reach the string =L/v
=0.40/8 s = 0.05 s
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
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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