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WAVE MOTION AND WAVES ON A STRING
EXERCISES:- Q-21 to Q-30
21. Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB (Figure 15-E5). The linear mass density of the wire AB is 10 g/m and that of CD is 8 g/m. Find the speed of a transverse wave pulse produced in AB and CD.
Figure for Q-21
ANSWER: The tension in the wire AB, F =2*3.2*9.8 N =62.72 N
The linear mass density of the wire AB, µ = 10 g/m =0.01 kg/m.
The wave speed in the wire AB,
v =√(F/µ) =√(62.72/0.01) =√6272 =79 m/s
The tension in the wire CD, F' =3.2*9.8 N =31.36 N
The linear mass density of the wire CD, µ' =8 g/m =0.008 kg/m.
The wave speed in the wire CD,
v' = √(F'/µ') =√(31.36/0.008) =√3920 =62.61 m/s ≈63 m/s
22. In the arrangement shown in figure (15-E6), the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m/s².
Figure for Q-22
ANSWER: The length of the string = 2.0 + 0.25 m =2.25 m
The mass of the string = 4.5 g. The linear mass of the string,
µ= 4.5/(2.25*1000) = 0.002 kg/m.
The tension in the string, F = 2*9.8 =19.6 N
The wave speed, v =√(F/µ) =√(19.6/0.002) =√9800 =99 m/s
The distance between floor to the pulley, L = 2.0 m
Hence time taken by the transverse disturbance to reach pulley from the floor = L/v =2.0/99 s =0.02 s
23. A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of 19.2x10⁻³ kg/m. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2.0 m/s². Take g = 10 m/s².
ANSWER: In this case the tension in the string, F =m(g+a)
=4.0*(10+2) N =48 N.
The linear mass density, µ = 19.2x10⁻³ kg/m
The wave speed, v = √(F/µ) =√(48/19.2x10⁻³) =√2500 =50 m/s
24. A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm/s on the string when the car is at rest and 62 cm/s when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m/s².
ANSWER: Let the mass of the ball = m. The tension in the string when the car is at rest, F = mg. Wave speed in this state v = 60 cm/s
→v =0.60 m/s. If the linear mass density of the string be µ,
√(F/µ) = v
→µ =F/v².
In the second case let the acceleration of the car = a. The pseudo force on the ball, P = ma. The tension in the string, F' = The resultant force on the ball =√(F²+P²)
Diagram for Q-24
The wave speed on the string, v' =62 cm/s =0.62 m/s
v' = √(F'/µ) =√(F'v²/F) =√{v²√(F²+P²)/F}
→v'² = v²√(1+P²/F²) =v²√(1+a²/g²)
→√(1+a²/g²) =v'²/v²
→1+a²/g² =(v'²/v²)²
→a²/g² = (v'²/v²)² - 1 =(0.62/60)⁴- 1 =1.14 -1 =0.14
→a² = 0.14g² =0.14*10² =14
→a = 3.7 m/s²
25. A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is v. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to string) will this disturbance travel on the string?
ANSWER: Let the radius of the ring =r and the mass per unit length µ. The speed of the ring =v. Outward force per unit length P =µv²/r. If the tension in the string = F, then considering a semicircular part of the ring,
2F = P*2r
→F =Pr =(µv²/r)*r =µv²
The speed of the transverse disturbance V =√(F/µ)
→V = √(µv²/µ) = √v² = v
26. A heavy but uniform rope of length L is suspended from a ceiling.
(a) Write the velocity of a transverse wave traveling on the string as a function of the distance from the lower end.
(b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling?
(c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse?
ANSWER: (a) Let the mass per unit length of the string = µ. If we consider any point on the rope at a distance x from the lower end, then the tension at this point F =µxg.
The velocity of a transverse wave at this point v =√(F/µ)
→v = √(µxg/µ)
→v = √(gx)
(b) At a distance x from the lower end, the infinitesimal time dt taken to travel an infinitesimal distance dx is
dt = dx/v =dx/√(gx)
To get the total time taken, T by the pulse to reach the ceiling from the bottom, we integrate dt from t=0 to t = T and the corresponding distance x = 0 to x = L.
T =∫dt =∫{1/√(gx)}dx =(1/√g)∫(1/√x)dx
→T =(1/√g)[2√x]
→T = (2/√g)[√L] {After putting the limits}
→T = 2√L/√g =√(4L/g)
(c) Let the pulse meet at a distance D from the lower end with the falling particle. The particle falls a distance of L-D. If the time taken by the particle to travel this distance = t.
L-D =½gt²
→t² = 2(L-D)/g
Same time will be taken by the wave to travel the distance D. From the expression in (b),
t = √(4D/g)
→t² = 4D/g
Equating both the values of t²,
2(L-D)/g = 4D/g
→2L-2D =4D
→6D = 2L
→D = L/3
Hence the wave and the particle will meet at a distance L/3 from the bottom.
27. Two long strings A and B, each having linear mass density 1.2x10⁻² kg/m, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x =0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?
ANSWER: The linear mass density of the strings µ = 1.2 x 10⁻² kg/m. The tension in A =4.8 N, tension in B =7.5 N. The wave speed in A, v =√(4.8/µ) =√400 = 20 m/s.
and the wave speed in B, v' = √(7.5/µ) =√625 = 25 m/s.
If the pulse on B overtakes the pulse on A at a distance x from the left end, the time taken t by the pulse on A to travel the x distance
t = x/20.
The same distance is traveled by the pulse on B in time t' = x/25. But this pulse is produced after a time 20 ms = 0.020 s. Hence,
t = t' + 0.02
→x/20 = 0.02 + x/25
→{(5-4)/100} x =0.02
→x = 0.02*100 = 2.0 m
And t = x/20 = 2/20 s =0.10 s =100 ms
28. A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the source transmitting to the wire?
ANSWER: A = 0.50 mm =5 x10⁻⁴ m, ν = 100 Hz, F = 100 N, v = 100 m/s.
⍵ = 2πν
The average power being transmitted by the source to the wire is given by
Pₐᵥ = ⍵²A²F/2v = 2π²ν²A²F/2v
=4π²*100²*(5 x10⁻⁴)²*100/200
=50π²*10⁻⁴ W
≈490*10⁻⁴ W
=49 mW
29. A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N.
(a) Find the average power transmitted across a given point on the string.
(b) Find the total energy associated with the wave in a 2.0 m long portion of the string.
ANSWER: (a) Frequency, ν = 200 Hz,
Amplitude, A = 1mm =1x10⁻³ m
Linear mass density µ = 6 g/m =0.006 kg/m
Tension, F = 60 N
Wave speed, v =√(F/µ) =√(60/0.006) =√10000 =100 m/s
Average power transmitted, Pₐᵥ = 2π²µvA²ν²
=2π²*0.006*100*(1x10⁻³)²*200² W
=0.47 W
(b) Time taken by the wave to travel a distance of 2.0 m
t = 2.0/100 s =0.02 s
The total energy associated in this time with a power 0.47 W
=0.47*0.02 J
=0.0094 J
=9.4 mJ
30. A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0.01 kg/m kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string.
(a) Find the wave speed and the wavelength of the waves.
(b) Find the maximum speed and acceleration of a particle of the string.
(c) At what average rate is the tuning fork transmitting energy to the string?
ANSWER: The tension in the string, F = 49 N. Linear mass density, µ = 0.01 kg/m. The amplitude of the waves, A = 0.50 mm =5 x10⁻⁴ m. The frequency of the wave, ν = 440 Hz.
(a) The wave speed, v =√(F/µ) =√(49/0.01) =√4900 =70 m/s.
The wavelength, 𝜆 =v/ν =70/440 m =7*100/44 cm
→𝜆 =15.9 cm ≈16 cm
(b) The magnitude of the maximum acceleration of a particle will be, a =⍵²A
{At the extreme of the displacement as in SHM}
→a = (2πν)²A
→a = 4π²*440²*5 x10⁻⁴ m/s
→a = 3.8 m/s²
The speed of a particle is given as V =⍵√(A²-x²)
Where x is the displacement from the mean position. Obviously, the maximum speed is at x = 0 i.e. at the mean position. Therefore,
Vₘₐₓ = ⍵A =2πνA =2π*440*5 x10⁻⁴ m/s =1.38 m/s ≈1.4 m/s
(c) The average power transmitted by the tuning fork,
Pₐᵥ =2π²µvA²ν²
=2π²*0.01*70*(5 x10⁻⁴)²*440² W
=35π²*44²*10⁻⁶ W
=0.67 W
21. Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB (Figure 15-E5). The linear mass density of the wire AB is 10 g/m and that of CD is 8 g/m. Find the speed of a transverse wave pulse produced in AB and CD.
ANSWER: The tension in the wire AB, F =2*3.2*9.8 N =62.72 N
The linear mass density of the wire AB, µ = 10 g/m =0.01 kg/m.
The wave speed in the wire AB,
v =√(F/µ) =√(62.72/0.01) =√6272 =79 m/s
The tension in the wire CD, F' =3.2*9.8 N =31.36 N
The linear mass density of the wire CD, µ' =8 g/m =0.008 kg/m.
The wave speed in the wire CD,
v' = √(F'/µ') =√(31.36/0.008) =√3920 =62.61 m/s ≈63 m/s
22. In the arrangement shown in figure (15-E6), the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m/s².
ANSWER: The length of the string = 2.0 + 0.25 m =2.25 m
The mass of the string = 4.5 g. The linear mass of the string,
µ= 4.5/(2.25*1000) = 0.002 kg/m.
The tension in the string, F = 2*9.8 =19.6 N
The wave speed, v =√(F/µ) =√(19.6/0.002) =√9800 =99 m/s
The distance between floor to the pulley, L = 2.0 m
Hence time taken by the transverse disturbance to reach pulley from the floor = L/v =2.0/99 s =0.02 s
23. A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of 19.2x10⁻³ kg/m. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2.0 m/s². Take g = 10 m/s².
ANSWER: In this case the tension in the string, F =m(g+a)
=4.0*(10+2) N =48 N.
The linear mass density, µ = 19.2x10⁻³ kg/m
The wave speed, v = √(F/µ) =√(48/19.2x10⁻³) =√2500 =50 m/s
24. A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm/s on the string when the car is at rest and 62 cm/s when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m/s².
ANSWER: Let the mass of the ball = m. The tension in the string when the car is at rest, F = mg. Wave speed in this state v = 60 cm/s
→v =0.60 m/s. If the linear mass density of the string be µ,
√(F/µ) = v
→µ =F/v².
In the second case let the acceleration of the car = a. The pseudo force on the ball, P = ma. The tension in the string, F' = The resultant force on the ball =√(F²+P²)
The wave speed on the string, v' =62 cm/s =0.62 m/s
v' = √(F'/µ) =√(F'v²/F) =√{v²√(F²+P²)/F}
→v'² = v²√(1+P²/F²) =v²√(1+a²/g²)
→√(1+a²/g²) =v'²/v²
→1+a²/g² =(v'²/v²)²
→a²/g² = (v'²/v²)² - 1 =(0.62/60)⁴- 1 =1.14 -1 =0.14
→a² = 0.14g² =0.14*10² =14
→a = 3.7 m/s²
25. A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is v. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to string) will this disturbance travel on the string?
ANSWER: Let the radius of the ring =r and the mass per unit length µ. The speed of the ring =v. Outward force per unit length P =µv²/r. If the tension in the string = F, then considering a semicircular part of the ring,
2F = P*2r
→F =Pr =(µv²/r)*r =µv²
The speed of the transverse disturbance V =√(F/µ)
→V = √(µv²/µ) = √v² = v
26. A heavy but uniform rope of length L is suspended from a ceiling.
(a) Write the velocity of a transverse wave traveling on the string as a function of the distance from the lower end.
(b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling?
(c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse?
ANSWER: (a) Let the mass per unit length of the string = µ. If we consider any point on the rope at a distance x from the lower end, then the tension at this point F =µxg.
The velocity of a transverse wave at this point v =√(F/µ)
→v = √(µxg/µ)
→v = √(gx)
(b) At a distance x from the lower end, the infinitesimal time dt taken to travel an infinitesimal distance dx is
dt = dx/v =dx/√(gx)
To get the total time taken, T by the pulse to reach the ceiling from the bottom, we integrate dt from t=0 to t = T and the corresponding distance x = 0 to x = L.
T =∫dt =∫{1/√(gx)}dx =(1/√g)∫(1/√x)dx
→T =(1/√g)[2√x]
→T = (2/√g)[√L] {After putting the limits}
→T = 2√L/√g =√(4L/g)
(c) Let the pulse meet at a distance D from the lower end with the falling particle. The particle falls a distance of L-D. If the time taken by the particle to travel this distance = t.
L-D =½gt²
→t² = 2(L-D)/g
Same time will be taken by the wave to travel the distance D. From the expression in (b),
t = √(4D/g)
→t² = 4D/g
Equating both the values of t²,
2(L-D)/g = 4D/g
→2L-2D =4D
→6D = 2L
→D = L/3
Hence the wave and the particle will meet at a distance L/3 from the bottom.
27. Two long strings A and B, each having linear mass density 1.2x10⁻² kg/m, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x =0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?
ANSWER: The linear mass density of the strings µ = 1.2 x 10⁻² kg/m. The tension in A =4.8 N, tension in B =7.5 N. The wave speed in A, v =√(4.8/µ) =√400 = 20 m/s.
and the wave speed in B, v' = √(7.5/µ) =√625 = 25 m/s.
If the pulse on B overtakes the pulse on A at a distance x from the left end, the time taken t by the pulse on A to travel the x distance
t = x/20.
The same distance is traveled by the pulse on B in time t' = x/25. But this pulse is produced after a time 20 ms = 0.020 s. Hence,
t = t' + 0.02
→x/20 = 0.02 + x/25
→{(5-4)/100} x =0.02
→x = 0.02*100 = 2.0 m
And t = x/20 = 2/20 s =0.10 s =100 ms
28. A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the source transmitting to the wire?
ANSWER: A = 0.50 mm =5 x10⁻⁴ m, ν = 100 Hz, F = 100 N, v = 100 m/s.
⍵ = 2πν
The average power being transmitted by the source to the wire is given by
Pₐᵥ = ⍵²A²F/2v = 2π²ν²A²F/2v
=4π²*100²*(5 x10⁻⁴)²*100/200
=50π²*10⁻⁴ W
≈490*10⁻⁴ W
=49 mW
29. A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N.
(a) Find the average power transmitted across a given point on the string.
(b) Find the total energy associated with the wave in a 2.0 m long portion of the string.
ANSWER: (a) Frequency, ν = 200 Hz,
Amplitude, A = 1mm =1x10⁻³ m
Linear mass density µ = 6 g/m =0.006 kg/m
Tension, F = 60 N
Wave speed, v =√(F/µ) =√(60/0.006) =√10000 =100 m/s
Average power transmitted, Pₐᵥ = 2π²µvA²ν²
=2π²*0.006*100*(1x10⁻³)²*200² W
=0.47 W
(b) Time taken by the wave to travel a distance of 2.0 m
t = 2.0/100 s =0.02 s
The total energy associated in this time with a power 0.47 W
=0.47*0.02 J
=0.0094 J
=9.4 mJ
30. A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0.01 kg/m kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string.
(a) Find the wave speed and the wavelength of the waves.
(b) Find the maximum speed and acceleration of a particle of the string.
(c) At what average rate is the tuning fork transmitting energy to the string?
ANSWER: The tension in the string, F = 49 N. Linear mass density, µ = 0.01 kg/m. The amplitude of the waves, A = 0.50 mm =5 x10⁻⁴ m. The frequency of the wave, ν = 440 Hz.
(a) The wave speed, v =√(F/µ) =√(49/0.01) =√4900 =70 m/s.
The wavelength, 𝜆 =v/ν =70/440 m =7*100/44 cm
→𝜆 =15.9 cm ≈16 cm
(b) The magnitude of the maximum acceleration of a particle will be, a =⍵²A
{At the extreme of the displacement as in SHM}
→a = (2πν)²A
→a = 4π²*440²*5 x10⁻⁴ m/s
→a = 3.8 m/s²
The speed of a particle is given as V =⍵√(A²-x²)
Where x is the displacement from the mean position. Obviously, the maximum speed is at x = 0 i.e. at the mean position. Therefore,
Vₘₐₓ = ⍵A =2πνA =2π*440*5 x10⁻⁴ m/s =1.38 m/s ≈1.4 m/s
(c) The average power transmitted by the tuning fork,
Pₐᵥ =2π²µvA²ν²
=2π²*0.01*70*(5 x10⁻⁴)²*440² W
=35π²*44²*10⁻⁶ W
=0.67 W
|
| Figure for Q-21 |
ANSWER: The tension in the wire AB, F =2*3.2*9.8 N =62.72 N
The linear mass density of the wire AB, µ = 10 g/m =0.01 kg/m.
The wave speed in the wire AB,
v =√(F/µ) =√(62.72/0.01) =√6272 =79 m/s
The tension in the wire CD, F' =3.2*9.8 N =31.36 N
The linear mass density of the wire CD, µ' =8 g/m =0.008 kg/m.
The wave speed in the wire CD,
v' = √(F'/µ') =√(31.36/0.008) =√3920 =62.61 m/s ≈63 m/s
22. In the arrangement shown in figure (15-E6), the string has a mass of 4.5 g. How much time will it take for a transverse disturbance produced at the floor to reach the pulley? Take g = 10 m/s².
|
| Figure for Q-22 |
ANSWER: The length of the string = 2.0 + 0.25 m =2.25 m
The mass of the string = 4.5 g. The linear mass of the string,
µ= 4.5/(2.25*1000) = 0.002 kg/m.
The tension in the string, F = 2*9.8 =19.6 N
The wave speed, v =√(F/µ) =√(19.6/0.002) =√9800 =99 m/s
The distance between floor to the pulley, L = 2.0 m
Hence time taken by the transverse disturbance to reach pulley from the floor = L/v =2.0/99 s =0.02 s
23. A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of 19.2x10⁻³ kg/m. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2.0 m/s². Take g = 10 m/s².
ANSWER: In this case the tension in the string, F =m(g+a)
=4.0*(10+2) N =48 N.
The linear mass density, µ = 19.2x10⁻³ kg/m
The wave speed, v = √(F/µ) =√(48/19.2x10⁻³) =√2500 =50 m/s
24. A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of 60 cm/s on the string when the car is at rest and 62 cm/s when the car accelerates on a horizontal road. Find the acceleration of the car. Take g = 10 m/s².
ANSWER: Let the mass of the ball = m. The tension in the string when the car is at rest, F = mg. Wave speed in this state v = 60 cm/s
→v =0.60 m/s. If the linear mass density of the string be µ,
√(F/µ) = v
→µ =F/v².
In the second case let the acceleration of the car = a. The pseudo force on the ball, P = ma. The tension in the string, F' = The resultant force on the ball =√(F²+P²)
 |
| Diagram for Q-24 |
The wave speed on the string, v' =62 cm/s =0.62 m/s
v' = √(F'/µ) =√(F'v²/F) =√{v²√(F²+P²)/F}
→v'² = v²√(1+P²/F²) =v²√(1+a²/g²)
→√(1+a²/g²) =v'²/v²
→1+a²/g² =(v'²/v²)²
→a²/g² = (v'²/v²)² - 1 =(0.62/60)⁴- 1 =1.14 -1 =0.14
→a² = 0.14g² =0.14*10² =14
→a = 3.7 m/s²
25. A circular loop of string rotates about its axis on a frictionless horizontal plane at a uniform rate so that the tangential speed of any particle of the string is v. If a small transverse disturbance is produced at a point of the loop, with what speed (relative to string) will this disturbance travel on the string?
ANSWER: Let the radius of the ring =r and the mass per unit length µ. The speed of the ring =v. Outward force per unit length P =µv²/r. If the tension in the string = F, then considering a semicircular part of the ring,
2F = P*2r
→F =Pr =(µv²/r)*r =µv²
The speed of the transverse disturbance V =√(F/µ)
→V = √(µv²/µ) = √v² = v
26. A heavy but uniform rope of length L is suspended from a ceiling.
(a) Write the velocity of a transverse wave traveling on the string as a function of the distance from the lower end.
(b) If the rope is given a sudden sideways jerk at the bottom, how long will it take for the pulse to reach the ceiling?
(c) A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse?
ANSWER: (a) Let the mass per unit length of the string = µ. If we consider any point on the rope at a distance x from the lower end, then the tension at this point F =µxg.
The velocity of a transverse wave at this point v =√(F/µ)
→v = √(µxg/µ)
→v = √(gx)
(b) At a distance x from the lower end, the infinitesimal time dt taken to travel an infinitesimal distance dx is
dt = dx/v =dx/√(gx)
To get the total time taken, T by the pulse to reach the ceiling from the bottom, we integrate dt from t=0 to t = T and the corresponding distance x = 0 to x = L.
T =∫dt =∫{1/√(gx)}dx =(1/√g)∫(1/√x)dx
→T =(1/√g)[2√x]
→T = (2/√g)[√L] {After putting the limits}
→T = 2√L/√g =√(4L/g)
(c) Let the pulse meet at a distance D from the lower end with the falling particle. The particle falls a distance of L-D. If the time taken by the particle to travel this distance = t.
L-D =½gt²
→t² = 2(L-D)/g
Same time will be taken by the wave to travel the distance D. From the expression in (b),
t = √(4D/g)
→t² = 4D/g
Equating both the values of t²,
2(L-D)/g = 4D/g
→2L-2D =4D
→6D = 2L
→D = L/3
Hence the wave and the particle will meet at a distance L/3 from the bottom.
27. Two long strings A and B, each having linear mass density 1.2x10⁻² kg/m, are stretched by different tensions 4.8 N and 7.5 N respectively and are kept parallel to each other with their left ends at x =0. Wave pulses are produced on the strings at the left ends at t = 0 on string A and at t = 20 ms on string B. When and where will the pulse on B overtake that on A?
ANSWER: The linear mass density of the strings µ = 1.2 x 10⁻² kg/m. The tension in A =4.8 N, tension in B =7.5 N. The wave speed in A, v =√(4.8/µ) =√400 = 20 m/s.
and the wave speed in B, v' = √(7.5/µ) =√625 = 25 m/s.
If the pulse on B overtakes the pulse on A at a distance x from the left end, the time taken t by the pulse on A to travel the x distance
t = x/20.
The same distance is traveled by the pulse on B in time t' = x/25. But this pulse is produced after a time 20 ms = 0.020 s. Hence,
t = t' + 0.02
→x/20 = 0.02 + x/25
→{(5-4)/100} x =0.02
→x = 0.02*100 = 2.0 m
And t = x/20 = 2/20 s =0.10 s =100 ms
28. A transverse wave of amplitude 0.50 mm and frequency 100 Hz is produced on a wire stretched to a tension of 100 N. If the wave speed is 100 m/s, what average power is the source transmitting to the wire?
ANSWER: A = 0.50 mm =5 x10⁻⁴ m, ν = 100 Hz, F = 100 N, v = 100 m/s.
⍵ = 2πν
The average power being transmitted by the source to the wire is given by
Pₐᵥ = ⍵²A²F/2v = 2π²ν²A²F/2v
=4π²*100²*(5 x10⁻⁴)²*100/200
=50π²*10⁻⁴ W
≈490*10⁻⁴ W
=49 mW
29. A 200 Hz wave with amplitude 1 mm travels on a long string of linear mass density 6 g/m kept under a tension of 60 N.
(a) Find the average power transmitted across a given point on the string.
(b) Find the total energy associated with the wave in a 2.0 m long portion of the string.
ANSWER: (a) Frequency, ν = 200 Hz,
Amplitude, A = 1mm =1x10⁻³ m
Linear mass density µ = 6 g/m =0.006 kg/m
Tension, F = 60 N
Wave speed, v =√(F/µ) =√(60/0.006) =√10000 =100 m/s
Average power transmitted, Pₐᵥ = 2π²µvA²ν²
=2π²*0.006*100*(1x10⁻³)²*200² W
=0.47 W
(b) Time taken by the wave to travel a distance of 2.0 m
t = 2.0/100 s =0.02 s
The total energy associated in this time with a power 0.47 W
=0.47*0.02 J
=0.0094 J
=9.4 mJ
30. A tuning fork of frequency 440 Hz is attached to a long string of linear mass density 0.01 kg/m kept under a tension of 49 N. The fork produces transverse waves of amplitude 0.50 mm on the string.
(a) Find the wave speed and the wavelength of the waves.
(b) Find the maximum speed and acceleration of a particle of the string.
(c) At what average rate is the tuning fork transmitting energy to the string?
ANSWER: The tension in the string, F = 49 N. Linear mass density, µ = 0.01 kg/m. The amplitude of the waves, A = 0.50 mm =5 x10⁻⁴ m. The frequency of the wave, ν = 440 Hz.
(a) The wave speed, v =√(F/µ) =√(49/0.01) =√4900 =70 m/s.
The wavelength, 𝜆 =v/ν =70/440 m =7*100/44 cm
→𝜆 =15.9 cm ≈16 cm
(b) The magnitude of the maximum acceleration of a particle will be, a =⍵²A
{At the extreme of the displacement as in SHM}
→a = (2πν)²A
→a = 4π²*440²*5 x10⁻⁴ m/s
→a = 3.8 m/s²
The speed of a particle is given as V =⍵√(A²-x²)
Where x is the displacement from the mean position. Obviously, the maximum speed is at x = 0 i.e. at the mean position. Therefore,
Vₘₐₓ = ⍵A =2πνA =2π*440*5 x10⁻⁴ m/s =1.38 m/s ≈1.4 m/s
(c) The average power transmitted by the tuning fork,
Pₐᵥ =2π²µvA²ν²
=2π²*0.01*70*(5 x10⁻⁴)²*440² W
=35π²*44²*10⁻⁶ W
=0.67 W
===<<<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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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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Click here for → QUESTIONS FOR SHORT ANSWER
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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"
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Click here for "The Forces" - Exercises
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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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