My Channel on YouTube → SimplePhysics with KK
For links to
other chapters - See bottom of the page
Or click here → kktutor.blogspot.com
WAVE MOTION AND WAVES ON A STRING
EXERCISES:- Q-1 to Q-10
1. A wave pulse passing on a string with a speed of 40 cm/s in the negative x-direction has its maximum at x = 0 at t = 0. Where will this maximum be located at t = 5 s?
ANSWER: Since the speed of the wave pulse is 40 cm/s, the maximum of the pulse will also travel with a speed of 40 cm/s in the negative x-direction. In a time of 5 s it will be at (-40 cm/s)*(5 s) = -200 cm = -2.0 m
2. The equation of a wave traveling on a string stretched along the x-axis is given by
y = A e–(x/a+t/T)²
(a) Write the dimensions of A, a and T.
(b) Find the wave speed.
(c) In which direction is the wave traveling?
(d) Where is the maximum of the pulse located at t = T? At t = 2T?
ANSWER: (a) Since y is the displacement of a particle on the string, its dimensions are [M⁰L¹T⁰]. The dimensions on the right-hand side of the expression must also be the same. 'A' is the amplitude of the wave, its dimensions will also be the same i.e. [M⁰L¹T⁰] or [L].
'e' is a number, so dimensionless. Therefore the expression to be dimensionally correct, each term of the expression x/a+t/T should also be dimensionless. Now, the 'x' is length, hence 'a' should also have the dimension of a length.
Hence the dimensions of 'a' = [M⁰L¹T⁰] or [L]
Similarly 't' is time, the dimensions of t = [M⁰L⁰T¹] or [T], hence same will be the dimensions of T = [M⁰L⁰T¹] or [T].
(b) The particle displacement in a wave equation is a function of (t-x/v), i.e. y = ƒ(t-x/v) -------- (i)
It is for the wave moving in the positive x-direction.
The given equation can also be written as,
y = A e–(x/a+t/T)²
→y = A e-{t+x/(a/T)}²/T² =ƒ(t+x/v)
Comparing we find wave speed v = a/T
(c) Since the sign of 'x' is opposite to equation (i), the wave is traveling in the negative x-direction.
(d) At t =0 and x =0, y = A e⁰ = A
It means initially at t = 0, the maximum of the pulse is located at x = 0 i.e. at the origin.
The velocity of the wave pulse v = -a/T, hence the maximum of the pulse at t = T will be at (-a/T)*T = -a.
And the maximum of the pulse at t = 2T will be at (-a/T)*2T = -2a.
3. Figure (15-E1) shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10 cm/s. Sketch the shape of the string at t = 1 s, 2 s and 3 s.
Figure for Q-3
ANSWER: Given that v = 10 cm/s. From the graph, at t = 0 the left end of the pulse is at x = 5 cm. At t = 1 s, 2 s, 3s the left end of the pulse will move to x= (5 + 10*1) cm, (5 + 10*2) cm, (5 + 10*3) cm respectively. i.e. to x = 15 cm, 25 cm, 35 cm respectively. The separation between the left end and the right end is 5 cm from the graph. Hence following will be the sketch of the pulse at t = 1 s, 2 s and 3 s.
Sketch for Q - 3
4. A pulse traveling on a string is represented by the function
y = a³/{(x-vt)² + a²}
where a = 5 mm and v = 20 cm/s. Sketch the shape of the string at t = 0, 1 s and 2 s. Take x = 0 in the middle of the string.
ANSWER: The given equation, y = a[{1+(x-vt)²/a²}–¹]
→y ≈ a{1-(x-vt)²/a²}, expanding and neglecting the third and higher terms.
→y = a - (x-vt)²/a
At t = 0, y = a - x²/a. So the shape is parabolic.
Putting y = 0, x² = a² → x = 土a
At x = 0, y = a. So at t = 0, the shape of the pulse will be as below:-
Sketch of the pulse at t = 0
At t = 1 s, y = a - (x - 20)²/a
If y = 0, (x-20)² = a² →x-20 = 土a, →x = 20 土 a =20土0.5 cm
→x = 20.5 cm and 19.5 cm.
At x = 20 cm, y = a = 0.5 cm
Hence at t = 1 s, the shape of the pulse will be as below:-
Sketch of the pulse at t = 1 s
Similarly at t = 2 s, y = a - (x-40)²/a
If y = 0, x = 40土a = 40土0.5 cm
→x = 39.5 cm and 40.5 cm.
At x =40 cm, y = a = 0.5 cm
So at t = 2 s, the shape of the pulse will be as below:-
Sketch of the pulse at t = 2s
5. The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive in the positive x-direction is given by f(t) = A sin (t/T). The wave speed is v. Write the wave equation.
ANSWER: The wave speed is v. At any time t, f(t) = A sin(t/T). Suppose in next time t' the distance traveled is x. Hence time-taken (t') in traveling a distance x = x/v. But the particle at distance x, has the same displacement as at time t-t'. Thus
f(x,t) = A sin {(t-t')/T)}= A sin {(t/T-t'/T)}
→f(x,t) = A sin {t/T - x/(vT)}
6. A wave pulse is traveling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by
g(x) = A sin(x/a), where A and a are constants.
(a) What are the dimensions of A and a? (b) Write the equation of the wave for a general time t, if the wave speed is v.
ANSWER: (a) The shape of the string at t = 0 is the displacement of the particle in the y-direction. Hence g(x) has a unit of length. On the right-hand side, there are two factors, A and sine. Since sine is a ratio, it is dimensionless. Hence the expression will be dimensionally correct only if A has also the unit of length. So the dimension of A = [M⁰L¹T⁰] or [L]
The angle of sine x/a is in radian which is again dimensionless. The x/a to be dimensionless, both of x and a should have the same unit. Since x is the distance, hence the unit of 'a' should also be the unit of distance. So the dimension of 'a' = [M⁰L¹T⁰] or [L]
(b) Given g(x) = A sin (x/a)
Consider the particle at further distance x' where the wave reaches in time t. The wave speed = v. Hence x' = vt. This time the particle in consideration has the same displacement as time t earlier, i.e. x' distance earlier. So for a general time t, the displacement of the particle is same as at x-x'.
So g(x,t) = A sin [(x-x')/a]
→g(x,t) = A sin {(x-vt)/a}
7. A wave propagates on a string in the positive x-direction at a velocity v. The shape of the string at t = tₒ is given by
g(x,tₒ) = A sin(x/a). Write the wave equation for a general time t.
ANSWER: As in the previous solution of problem number 6, the further distance x' is traveled in time t. Clearly, this time difference t here is t-tₒ. Putting t-tₒ in place of t, in the above solution we get,
g(x,t) = A sin{x-v(t-tₒ)}/a
8. The equation of a wave traveling on a string is
y = (0.10 mm) sin[(31.4 m–1)x + (314 s–1)t].
(a) In which direction does the wave travel?
(b) Find the wave speed, the wavelength and the frequency of the wave.
(c) What is the maximum displacement and the maximum speed of a portion of the string?
ANSWER: (a) Since the coefficient of x is positive, the wave travels in the negative x-direction.
(b) Comparing it with the wave equation
y = A sin⍵(t-x/v) =A sin (⍵t - ⍵x/v)
⍵ = 314 s–1
-⍵/v = 31.4 m–1
→v = -314/31.4 m/s = -10 m/s
Frequency ν = ⍵/2π =314/(2*π) =100/2 = 50 Hz
Time period T = 1/ν =1/50 s
So the wavelength = vT = 10*(1/50) m =0.2 m = 20 cm
(c) Comparing with the wave equation
y = A sin (⍵t - ⍵x/v)
A = amplitude = 0.10 mm, Since the maximum value of sine =1, hence the maximum displacement of a portion of a string = A = 0.10 mm
The speed V is given as A⍵.cos(⍵t - ⍵x/v)
→V = (0.10 mm)(314 s–1)cos[(31.4 m–1)x + (314 s–1)t]
Since the maximum value of cosine =1, hence the maximum speed of a portion of a string = V' = A⍵ = (0.10 mm)(314 s–1)
=31.4 mm/s = 3.14 cm/s
9. A wave travels along the positive x-direction with a speed of 20 m/s. The amplitude of the wave is 0.20 cm and the wavelength 2.0 cm. (a) Write a suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2.0 cm at time t = 0 according to the wave equation written? Can you get different values of this quantity if the wave equation is written in a different fashion?
ANSWER: (a) The wave equation is given as
y = A sin(⍵t-kx)
Where wave number k = ⍵/v
Given v = 20 m/s = 2000 cm/s, A = 0.20 cm, Wavelength λ = 2 cm.
But λ = 2π/k =2πv/⍵
→2 = 2π*2000/⍵
→⍵ = 2000π
And k = ⍵/v =2000π/2000 =π
Hence the wave equation is
y = (0.20 cm) sin[(2000π s–1)t - (π cm–1)x]
(b) At t = 0, and x = 2 cm
Displacement = (0.20 cm) sin[-2π] = (0.20 cm)*0 = 0 {zero}
Velocity of the particle V = A⍵ cos(⍵t-kx)
=(0.20 cm)(2000π s–1) cos[(2000π s–1)t - (π cm–1)x]
At t = 0, and x = 2 cm
V = (0.20 cm)(2000π s–1) cos[ - (π cm–1)(2 cm)]
=400π*cos[-2π] =400π cm/s = 4π m/s
This wave equation has been written assuming that the wave is a sine wave. If the same sine wave is written in a different fashion, then the values of displacement and velocity of the particle will not change. But, if the wave profile is different than this, then the wave equation will be written in different form. In that case, the values may differ.
10. A wave is described by the equation
y = (1.0 mm) sinπ{x/2.0 cm - t/0.01 s}.
(a) Find the time period and the wavelength.
(b) Write the equation for the velocity of the particles. Find the speed of the particle at x =1.0 cm at time t = 0.01 s.
(c) What are the speeds of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s?
(d) What are the speeds of the particles at x = 1.0 cm at t = 0.011, 0.012, and 0.013 s?
ANSWER: (a) For the given equation, ⍵ = π/0.01 s–1
The time period T = 2π/⍵ = 2π/(π/0.01) =2/100 s =(2/100)*1000 ms
→ T = 20 ms
Wave number k = π/2 cm–1
But wavelength λ=2π/k = 2π/(π/2) cm = 4 cm
(b) The velocity of the particle
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
[Differentiating the given equation with respect to time]
At x = 1 cm and t = 0.01 s
V = (0.01 cm)(100π) cosπ[0.5-1]
→V = (0.01 cm)(100π) cos(-π/2) = 0 {zero}
[Since cos(-π/2) = 0]
(c) The speed at t =0.01 s and x = 3 cm
=(0.01 cm)(100π) cosπ[1.5-1]
=(0.01 cm)(100π) cosπ/2 = 0 {zero}
[Since cos π/2 = 0]
At x = 5 cm
Speed = (0.01 cm)(100π) cosπ[2.5-1]
=(0.01 cm)(100π) cos(3π/2) = 0 {zero}
[Since cos 3π/2 = 0]
At x = 7 cm
Speed = (0.01 cm)(100π) cosπ[3.5-1]
=(0.01 cm)(100π) cos(5π/2) = 0 {zero}
[Since cos 5π/2 = 0]
(d) Particle speed at x =1 cm is
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
→V =(0.1 cm)(100π) cosπ{0.5 - t/0.01 s}
For t = 0.011 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.1}
→V =(0.1 cm)(100π) cos{-0.6π}
→V =(0.1 cm)(100π) cos108° = 10π*0.31 cm/s = 9.7 cm/s
For t = 0.012 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.2}
→V =(0.1 cm)(100π) cos{-0.7π}
→V =(0.1 cm)(100π) cos126° = 10π*0.58 cm/s = 18.2 cm/s
For t = 0.013 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.3}
→V =(0.1 cm)(100π) cos{-0.8π}
→V =(0.1 cm)(100π) cos144° = 10π*0.8 cm/s = 25.1 cm/s
1. A wave pulse passing on a string with a speed of 40 cm/s in the negative x-direction has its maximum at x = 0 at t = 0. Where will this maximum be located at t = 5 s?
(a) Write the dimensions of A, a and T.
ANSWER: Since the speed of the wave pulse is 40 cm/s, the maximum of the pulse will also travel with a speed of 40 cm/s in the negative x-direction. In a time of 5 s it will be at (-40 cm/s)*(5 s) = -200 cm = -2.0 m
2. The equation of a wave traveling on a string stretched along the x-axis is given by
y = A e–(x/a+t/T)²
(a) Write the dimensions of A, a and T.
(b) Find the wave speed.
(c) In which direction is the wave traveling?
(d) Where is the maximum of the pulse located at t = T? At t = 2T?
ANSWER: (a) Since y is the displacement of a particle on the string, its dimensions are [M⁰L¹T⁰]. The dimensions on the right-hand side of the expression must also be the same. 'A' is the amplitude of the wave, its dimensions will also be the same i.e. [M⁰L¹T⁰] or [L].
'e' is a number, so dimensionless. Therefore the expression to be dimensionally correct, each term of the expression x/a+t/T should also be dimensionless. Now, the 'x' is length, hence 'a' should also have the dimension of a length.
Hence the dimensions of 'a' = [M⁰L¹T⁰] or [L]
Similarly 't' is time, the dimensions of t = [M⁰L⁰T¹] or [T], hence same will be the dimensions of T = [M⁰L⁰T¹] or [T].
(b) The particle displacement in a wave equation is a function of (t-x/v), i.e. y = ƒ(t-x/v) -------- (i)
It is for the wave moving in the positive x-direction.
The given equation can also be written as,
y = A e–(x/a+t/T)²
→y = A e-{t+x/(a/T)}²/T² =ƒ(t+x/v)
Comparing we find wave speed v = a/T
(c) Since the sign of 'x' is opposite to equation (i), the wave is traveling in the negative x-direction.
(d) At t =0 and x =0, y = A e⁰ = A
It means initially at t = 0, the maximum of the pulse is located at x = 0 i.e. at the origin.
The velocity of the wave pulse v = -a/T, hence the maximum of the pulse at t = T will be at (-a/T)*T = -a.
And the maximum of the pulse at t = 2T will be at (-a/T)*2T = -2a.
3. Figure (15-E1) shows a wave pulse at t = 0. The pulse moves to the right with a speed of 10 cm/s. Sketch the shape of the string at t = 1 s, 2 s and 3 s.
 |
| Figure for Q-3 |
ANSWER: Given that v = 10 cm/s. From the graph, at t = 0 the left end of the pulse is at x = 5 cm. At t = 1 s, 2 s, 3s the left end of the pulse will move to x= (5 + 10*1) cm, (5 + 10*2) cm, (5 + 10*3) cm respectively. i.e. to x = 15 cm, 25 cm, 35 cm respectively. The separation between the left end and the right end is 5 cm from the graph. Hence following will be the sketch of the pulse at t = 1 s, 2 s and 3 s.
4. A pulse traveling on a string is represented by the function
y = a³/{(x-vt)² + a²}
 |
| Sketch for Q - 3 |
4. A pulse traveling on a string is represented by the function
y = a³/{(x-vt)² + a²}
where a = 5 mm and v = 20 cm/s. Sketch the shape of the string at t = 0, 1 s and 2 s. Take x = 0 in the middle of the string.
ANSWER: The given equation, y = a[{1+(x-vt)²/a²}–¹]
→y ≈ a{1-(x-vt)²/a²}, expanding and neglecting the third and higher terms.
→y = a - (x-vt)²/a
At t = 0, y = a - x²/a. So the shape is parabolic.
→y ≈ a{1-(x-vt)²/a²}, expanding and neglecting the third and higher terms.
→y = a - (x-vt)²/a
At t = 0, y = a - x²/a. So the shape is parabolic.
Putting y = 0, x² = a² → x = 土a
At x = 0, y = a. So at t = 0, the shape of the pulse will be as below:-
 |
| Sketch of the pulse at t = 0 |
At t = 1 s, y = a - (x - 20)²/a
If y = 0, (x-20)² = a² →x-20 = 土a, →x = 20 土 a =20土0.5 cm
→x = 20.5 cm and 19.5 cm.
At x = 20 cm, y = a = 0.5 cm
Hence at t = 1 s, the shape of the pulse will be as below:-
 |
| Sketch of the pulse at t = 1 s |
Similarly at t = 2 s, y = a - (x-40)²/a
If y = 0, x = 40土a = 40土0.5 cm
→x = 39.5 cm and 40.5 cm.
At x =40 cm, y = a = 0.5 cm
So at t = 2 s, the shape of the pulse will be as below:-
 |
| Sketch of the pulse at t = 2s |
5. The displacement of the particle at x = 0 of a stretched string carrying a wave in the positive in the positive x-direction is given by f(t) = A sin (t/T). The wave speed is v. Write the wave equation.
ANSWER: The wave speed is v. At any time t, f(t) = A sin(t/T). Suppose in next time t' the distance traveled is x. Hence time-taken (t') in traveling a distance x = x/v. But the particle at distance x, has the same displacement as at time t-t'. Thus
f(x,t) = A sin {(t-t')/T)}= A sin {(t/T-t'/T)}
→f(x,t) = A sin {t/T - x/(vT)}
f(x,t) = A sin {(t-t')/T)}= A sin {(t/T-t'/T)}
→f(x,t) = A sin {t/T - x/(vT)}
6. A wave pulse is traveling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by
g(x) = A sin(x/a), where A and a are constants.
(a) What are the dimensions of A and a? (b) Write the equation of the wave for a general time t, if the wave speed is v.
ANSWER: (a) The shape of the string at t = 0 is the displacement of the particle in the y-direction. Hence g(x) has a unit of length. On the right-hand side, there are two factors, A and sine. Since sine is a ratio, it is dimensionless. Hence the expression will be dimensionally correct only if A has also the unit of length. So the dimension of A = [M⁰L¹T⁰] or [L]
The angle of sine x/a is in radian which is again dimensionless. The x/a to be dimensionless, both of x and a should have the same unit. Since x is the distance, hence the unit of 'a' should also be the unit of distance. So the dimension of 'a' = [M⁰L¹T⁰] or [L]
(b) Given g(x) = A sin (x/a)
Consider the particle at further distance x' where the wave reaches in time t. The wave speed = v. Hence x' = vt. This time the particle in consideration has the same displacement as time t earlier, i.e. x' distance earlier. So for a general time t, the displacement of the particle is same as at x-x'.
So g(x,t) = A sin [(x-x')/a]
→g(x,t) = A sin {(x-vt)/a}
The angle of sine x/a is in radian which is again dimensionless. The x/a to be dimensionless, both of x and a should have the same unit. Since x is the distance, hence the unit of 'a' should also be the unit of distance. So the dimension of 'a' = [M⁰L¹T⁰] or [L]
(b) Given g(x) = A sin (x/a)
Consider the particle at further distance x' where the wave reaches in time t. The wave speed = v. Hence x' = vt. This time the particle in consideration has the same displacement as time t earlier, i.e. x' distance earlier. So for a general time t, the displacement of the particle is same as at x-x'.
So g(x,t) = A sin [(x-x')/a]
→g(x,t) = A sin {(x-vt)/a}
7. A wave propagates on a string in the positive x-direction at a velocity v. The shape of the string at t = tₒ is given by
g(x,tₒ) = A sin(x/a). Write the wave equation for a general time t.
ANSWER: As in the previous solution of problem number 6, the further distance x' is traveled in time t. Clearly, this time difference t here is t-tₒ. Putting t-tₒ in place of t, in the above solution we get,
g(x,t) = A sin{x-v(t-tₒ)}/a
8. The equation of a wave traveling on a string is
y = (0.10 mm) sin[(31.4 m–1)x + (314 s–1)t].
(a) In which direction does the wave travel?
(b) Find the wave speed, the wavelength and the frequency of the wave.
(c) What is the maximum displacement and the maximum speed of a portion of the string?
g(x,t) = A sin{x-v(t-tₒ)}/a
8. The equation of a wave traveling on a string is
y = (0.10 mm) sin[(31.4 m–1)x + (314 s–1)t].
(a) In which direction does the wave travel?
(b) Find the wave speed, the wavelength and the frequency of the wave.
(c) What is the maximum displacement and the maximum speed of a portion of the string?
ANSWER: (a) Since the coefficient of x is positive, the wave travels in the negative x-direction.
(b) Comparing it with the wave equation
y = A sin⍵(t-x/v) =A sin (⍵t - ⍵x/v)
⍵ = 314 s–1
-⍵/v = 31.4 m–1
→v = -314/31.4 m/s = -10 m/s
Frequency ν = ⍵/2π =314/(2*π) =100/2 = 50 Hz
Time period T = 1/ν =1/50 s
So the wavelength = vT = 10*(1/50) m =0.2 m = 20 cm
(c) Comparing with the wave equation
y = A sin (⍵t - ⍵x/v)
A = amplitude = 0.10 mm, Since the maximum value of sine =1, hence the maximum displacement of a portion of a string = A = 0.10 mm
The speed V is given as A⍵.cos(⍵t - ⍵x/v)
→V = (0.10 mm)(314 s–1)cos[(31.4 m–1)x + (314 s–1)t]
Since the maximum value of cosine =1, hence the maximum speed of a portion of a string = V' = A⍵ = (0.10 mm)(314 s–1)
=31.4 mm/s = 3.14 cm/s
(b) Comparing it with the wave equation
y = A sin⍵(t-x/v) =A sin (⍵t - ⍵x/v)
⍵ = 314 s–1
-⍵/v = 31.4 m–1
→v = -314/31.4 m/s = -10 m/s
Frequency ν = ⍵/2π =314/(2*π) =100/2 = 50 Hz
Time period T = 1/ν =1/50 s
So the wavelength = vT = 10*(1/50) m =0.2 m = 20 cm
(c) Comparing with the wave equation
y = A sin (⍵t - ⍵x/v)
A = amplitude = 0.10 mm, Since the maximum value of sine =1, hence the maximum displacement of a portion of a string = A = 0.10 mm
The speed V is given as A⍵.cos(⍵t - ⍵x/v)
→V = (0.10 mm)(314 s–1)cos[(31.4 m–1)x + (314 s–1)t]
Since the maximum value of cosine =1, hence the maximum speed of a portion of a string = V' = A⍵ = (0.10 mm)(314 s–1)
=31.4 mm/s = 3.14 cm/s
9. A wave travels along the positive x-direction with a speed of 20 m/s. The amplitude of the wave is 0.20 cm and the wavelength 2.0 cm. (a) Write a suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2.0 cm at time t = 0 according to the wave equation written? Can you get different values of this quantity if the wave equation is written in a different fashion?
ANSWER: (a) The wave equation is given as
y = A sin(⍵t-kx)
Where wave number k = ⍵/v
Given v = 20 m/s = 2000 cm/s, A = 0.20 cm, Wavelength λ = 2 cm.
But λ = 2π/k =2πv/⍵
→2 = 2π*2000/⍵
→⍵ = 2000π
And k = ⍵/v =2000π/2000 =π
Hence the wave equation is
V = (0.20 cm)(2000π s–1) cos[ - (π cm–1)(2 cm)]
=400π*cos[-2π] =400π cm/s = 4π m/s
This wave equation has been written assuming that the wave is a sine wave. If the same sine wave is written in a different fashion, then the values of displacement and velocity of the particle will not change. But, if the wave profile is different than this, then the wave equation will be written in different form. In that case, the values may differ.
y = A sin(⍵t-kx)
Where wave number k = ⍵/v
Given v = 20 m/s = 2000 cm/s, A = 0.20 cm, Wavelength λ = 2 cm.
But λ = 2π/k =2πv/⍵
→2 = 2π*2000/⍵
→⍵ = 2000π
And k = ⍵/v =2000π/2000 =π
Hence the wave equation is
y = (0.20 cm) sin[(2000π s–1)t - (π cm–1)x]
(b) At t = 0, and x = 2 cm
Displacement = (0.20 cm) sin[-2π] = (0.20 cm)*0 = 0 {zero}
Velocity of the particle V = A⍵ cos(⍵t-kx)
=(0.20 cm)(2000π s–1) cos[(2000π s–1)t - (π cm–1)x]
At t = 0, and x = 2 cmV = (0.20 cm)(2000π s–1) cos[ - (π cm–1)(2 cm)]
=400π*cos[-2π] =400π cm/s = 4π m/s
This wave equation has been written assuming that the wave is a sine wave. If the same sine wave is written in a different fashion, then the values of displacement and velocity of the particle will not change. But, if the wave profile is different than this, then the wave equation will be written in different form. In that case, the values may differ.
10. A wave is described by the equation
y = (1.0 mm) sinπ{x/2.0 cm - t/0.01 s}.
(a) Find the time period and the wavelength.
(b) Write the equation for the velocity of the particles. Find the speed of the particle at x =1.0 cm at time t = 0.01 s.
(c) What are the speeds of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s?
(d) What are the speeds of the particles at x = 1.0 cm at t = 0.011, 0.012, and 0.013 s?
ANSWER: (a) For the given equation, ⍵ = π/0.01 s–1
The time period T = 2π/⍵ = 2π/(π/0.01) =2/100 s =(2/100)*1000 ms
→ T = 20 ms
Wave number k = π/2 cm–1
But wavelength λ=2π/k = 2π/(π/2) cm = 4 cm
(b) The velocity of the particle
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
[Differentiating the given equation with respect to time]
At x = 1 cm and t = 0.01 s
V = (0.01 cm)(100π) cosπ[0.5-1]
→V = (0.01 cm)(100π) cos(-π/2) = 0 {zero}
[Since cos(-π/2) = 0]
(c) The speed at t =0.01 s and x = 3 cm
=(0.01 cm)(100π) cosπ[1.5-1]
=(0.01 cm)(100π) cosπ/2 = 0 {zero}
[Since cos π/2 = 0]
At x = 5 cm
Speed = (0.01 cm)(100π) cosπ[2.5-1]
=(0.01 cm)(100π) cos(3π/2) = 0 {zero}
[Since cos 3π/2 = 0]
At x = 7 cm
Speed = (0.01 cm)(100π) cosπ[3.5-1]
=(0.01 cm)(100π) cos(5π/2) = 0 {zero}
[Since cos 5π/2 = 0]
(d) Particle speed at x =1 cm is
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
The time period T = 2π/⍵ = 2π/(π/0.01) =2/100 s =(2/100)*1000 ms
→ T = 20 ms
Wave number k = π/2 cm–1
But wavelength λ=2π/k = 2π/(π/2) cm = 4 cm
(b) The velocity of the particle
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
[Differentiating the given equation with respect to time]
At x = 1 cm and t = 0.01 s
V = (0.01 cm)(100π) cosπ[0.5-1]
→V = (0.01 cm)(100π) cos(-π/2) = 0 {zero}
[Since cos(-π/2) = 0]
(c) The speed at t =0.01 s and x = 3 cm
=(0.01 cm)(100π) cosπ[1.5-1]
=(0.01 cm)(100π) cosπ/2 = 0 {zero}
[Since cos π/2 = 0]
At x = 5 cm
Speed = (0.01 cm)(100π) cosπ[2.5-1]
=(0.01 cm)(100π) cos(3π/2) = 0 {zero}
[Since cos 3π/2 = 0]
At x = 7 cm
Speed = (0.01 cm)(100π) cosπ[3.5-1]
=(0.01 cm)(100π) cos(5π/2) = 0 {zero}
[Since cos 5π/2 = 0]
(d) Particle speed at x =1 cm is
V =(1.0 mm)(100π) cosπ{x/2.0 cm - t/0.01 s}
→V =(0.1 cm)(100π) cosπ{0.5 - t/0.01 s}
For t = 0.011 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.1}
→V =(0.1 cm)(100π) cos{-0.6π}
→V =(0.1 cm)(100π) cos108° = 10π*0.31 cm/s = 9.7 cm/s
→V =(0.1 cm)(100π) cos{-0.6π}
→V =(0.1 cm)(100π) cos108° = 10π*0.31 cm/s = 9.7 cm/s
For t = 0.012 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.2}
→V =(0.1 cm)(100π) cos{-0.7π}
→V =(0.1 cm)(100π) cos126° = 10π*0.58 cm/s = 18.2 cm/s
→V =(0.1 cm)(100π) cos{-0.7π}
→V =(0.1 cm)(100π) cos126° = 10π*0.58 cm/s = 18.2 cm/s
For t = 0.013 s
V =(0.1 cm)(100π) cosπ{0.5 - 1.3}
→V =(0.1 cm)(100π) cos{-0.8π}
→V =(0.1 cm)(100π) cos144° = 10π*0.8 cm/s = 25.1 cm/s
→V =(0.1 cm)(100π) cos{-0.8π}
→V =(0.1 cm)(100π) cos144° = 10π*0.8 cm/s = 25.1 cm/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
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
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)
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
Click here for → Questions for Short Answer
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)
---------------------------------------------------------------------------------
---------------------------------------------------------------------------------
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)
-------------------------------------------------------------------------------
-------------------------------------------------------------------------------
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
--------------------------------------------------------------------------------------------------------------
--------------------------------------------------------------------------------------------------------------
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"
No comments:
Post a Comment