Assignment 9
Traveling Wave
The wavefunction of a harmonic wave on a string is y(x,t) = 0.004 sin(70.4x + 298.0t), where x and y are in and t is in . What is the velocity of the wave taking positive to be in the +x direction and negative to be in the -x direction.
-4.23 m/s
What is the wavelength of the traveling wave?
8.92×10-2 m
What is the period of this wave?
2.11×10-2 s
-4.23 m/s
What is the wavelength of the traveling wave?
8.92×10-2 m
What is the period of this wave?
2.11×10-2 s
Discussion
NEW
Speed
In the notes its states that v=λ/T
I have found λ and got the correct answer
I have found T and got the correct answer
If anyone can see what i am doing wrong please help
NEW
Re: Speed Anonymous 2
The velocity should be the wavelength divided by the
period.
If you are still getting it wrong check the sign of the
velocity.
From the standard periodic wave equation, if the
coefficient for x has a negative before it, the direction
would be positive and therefore the velocity is positive.
For mine, the coefficient of x was a positive number so it
is in the -x direction and the velocity would be negative.
Hope that helps.
NEW
Re: Re: Speed
Thanks
NEW
I tried using the equation v= f * wavelength and i found my
f and by solving for w=2pif and wavelength by using
k=2pi/wavelenth and i multiplied those together and im
getting it wrong can anyone explain why?
NEW
Where to start? Anonymous 4
Can someone please help me to understand where I start on
this? Thank you so much for your help!
NEW
Re: Speed
im having the same issue, v should be lambada/T but i keep
getting the incorrect answer despite having the correct answer
for parts B and C
if some one could help id appreciate it
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Re: Re: Speed
o nvm it was just negative
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Help Anonymous 6
Can someone please explain how to find the velocity! Thank you!
NEW
For the velocity try changing your signs around.
v = lamda/period, which is your second answer divided by
your third. units are m/s.
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Anonymous 8
how do you find period?
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Re:
divide part 2 answer by part one
Speed
In the notes its states that v=λ/T
I have found λ and got the correct answer
I have found T and got the correct answer
If anyone can see what i am doing wrong please help
NEW
Re: Speed Anonymous 2
The velocity should be the wavelength divided by the
period.
If you are still getting it wrong check the sign of the
velocity.
From the standard periodic wave equation, if the
coefficient for x has a negative before it, the direction
would be positive and therefore the velocity is positive.
For mine, the coefficient of x was a positive number so it
is in the -x direction and the velocity would be negative.
Hope that helps.
NEW
Re: Re: Speed
Thanks
NEW
I tried using the equation v= f * wavelength and i found my
f and by solving for w=2pif and wavelength by using
k=2pi/wavelenth and i multiplied those together and im
getting it wrong can anyone explain why?
NEW
Where to start? Anonymous 4
Can someone please help me to understand where I start on
this? Thank you so much for your help!
NEW
Re: Speed
im having the same issue, v should be lambada/T but i keep
getting the incorrect answer despite having the correct answer
for parts B and C
if some one could help id appreciate it
NEW
Re: Re: Speed
o nvm it was just negative
NEW
Help Anonymous 6
Can someone please explain how to find the velocity! Thank you!
NEW
For the velocity try changing your signs around.
v = lamda/period, which is your second answer divided by
your third. units are m/s.
NEW
Anonymous 8
how do you find period?
NEW
Re:
divide part 2 answer by part one
Correct Tension on a Guitar String
A particular guitar string is supposed to vibrate at 195 Hz, but it is measured to actually vibrate at 200 Hz. By what percent should the tension in the string be changed to get the frequency to the correct value?
-4.94
-4.94
Discussion
NEW
Some help
This is how I worked the problem. I only used the equation
v=sqrt(FT/mu)
then some sort of logic...
Since we are only talking about one particular string we can
ignore the mu in this problem since it will remain constant.
Remember mu is equal to mass/length which is the same for
the same exact string. I took the number in Hz to equal the
velocity (v). We want to find the tension (FT).
In order to find the Tension on the guitar string when it
vibrates at the correct frequency plug the Hz given in as
the v variable (ignore the mu) and solve for tension. Do
the same for the other number given.
v^2 = Fr
Now take the difference in the tensions:
(correct tension - current tension)
take this number and divide by the tension found for the
correct tension:
difference/correct tension
and multiply by 100 to give a percent difference.
Since the initial frequency was higher than the correct
frequency we would lessen the tension and therefore the
change would be negative.
Hope this makes sense.
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Anonymous 3
this is a good run through except that you have to take the
difference/actual, not the difference/goal
NEW
% sign
Be sure to just enter the number ONLY. I wasted all of my
tries by adding the % sign in my answer. I had the right
answer on my last try and got it wrong because the % sign
is not in the answer.
NEW
not working?
hey so none of this is working for me??
i did 198^2= 39204
201^2= 40401
and the difference is -1197
-1197/198= -.03*100 = 30.5%
(and no i didnt put % in the answer)
i dont know what im doing wrong???
NEW
Re: not working? Anonymous 6
maybe you forgot the negative sign in your answer?
NEW
Frequency f for a given guitar string is proportional to the
square root of the tension T.
F = k sqrt(T)
If F1 and T1 are the new frequency and tension:
F1 = k sqrt(T1)
F1 / F = sqrt(T1 / T)
T1 / T = (F1 / F)^2
= (200 / 205)^2
= 0.9518.
The tension should be reduced by:
100(1 - 0.09518)
= 4.82
make sure the answer is negative!
NEW
Kayla
makes sure to put your difference in squared tension over
your larger number of squared tension
in other words
1197/40401=.0296
not
1197/201
hope this helps
Some help
This is how I worked the problem. I only used the equation
v=sqrt(FT/mu)
then some sort of logic...
Since we are only talking about one particular string we can
ignore the mu in this problem since it will remain constant.
Remember mu is equal to mass/length which is the same for
the same exact string. I took the number in Hz to equal the
velocity (v). We want to find the tension (FT).
In order to find the Tension on the guitar string when it
vibrates at the correct frequency plug the Hz given in as
the v variable (ignore the mu) and solve for tension. Do
the same for the other number given.
v^2 = Fr
Now take the difference in the tensions:
(correct tension - current tension)
take this number and divide by the tension found for the
correct tension:
difference/correct tension
and multiply by 100 to give a percent difference.
Since the initial frequency was higher than the correct
frequency we would lessen the tension and therefore the
change would be negative.
Hope this makes sense.
NEW
Anonymous 3
this is a good run through except that you have to take the
difference/actual, not the difference/goal
NEW
% sign
Be sure to just enter the number ONLY. I wasted all of my
tries by adding the % sign in my answer. I had the right
answer on my last try and got it wrong because the % sign
is not in the answer.
NEW
not working?
hey so none of this is working for me??
i did 198^2= 39204
201^2= 40401
and the difference is -1197
-1197/198= -.03*100 = 30.5%
(and no i didnt put % in the answer)
i dont know what im doing wrong???
NEW
Re: not working? Anonymous 6
maybe you forgot the negative sign in your answer?
NEW
Frequency f for a given guitar string is proportional to the
square root of the tension T.
F = k sqrt(T)
If F1 and T1 are the new frequency and tension:
F1 = k sqrt(T1)
F1 / F = sqrt(T1 / T)
T1 / T = (F1 / F)^2
= (200 / 205)^2
= 0.9518.
The tension should be reduced by:
100(1 - 0.09518)
= 4.82
make sure the answer is negative!
NEW
Kayla
makes sure to put your difference in squared tension over
your larger number of squared tension
in other words
1197/40401=.0296
not
1197/201
hope this helps
Two springs are connected, one is a heavy spring and the other is a light spring. A pulse is travelling down spring A, towards a junction with spring B. Which of the following statements are correct/incorrect.
Correct: The only case in which there will be no reflected pulse is when the springs are identical.
Correct: If spring A is the heavy spring then the transmitted pulse will not be inverted.
Correct: If spring A is the light spring then the transmitted pulse will not be inverted.
Correct: If spring A is the heavy spring then the reflected pulse will not be inverted.
Correct: If spring A is the light spring then the reflected pulse will travel faster than transmitted pulse.
Correct: The only case in which there will be no reflected pulse is when the springs are identical.
Correct: If spring A is the heavy spring then the transmitted pulse will not be inverted.
Correct: If spring A is the light spring then the transmitted pulse will not be inverted.
Correct: If spring A is the heavy spring then the reflected pulse will not be inverted.
Correct: If spring A is the light spring then the reflected pulse will travel faster than transmitted pulse.
Discussion
NEW
Help
Not sure about this one. Any pointers?
NEW
? Anonymous 2
Help please?!
NEW
Incorrect: If spring A is the heavy spring then the
transmitted pulse will travel slower than reflected pulse.
Correct: The only case in which there will be no reflected
pulse is when the springs are identical.
Incorrect: If spring A is the light spring then the
transmitted pulse will be inverted.
Incorrect: If spring A is the light spring then the
reflected pulse will not be inverted.
Correct: If spring A is the heavy spring then the reflected
pulse will not be inverted.
NEW
Thanks!!! Anonymous 4
Thank you so much!!!
Help
Not sure about this one. Any pointers?
NEW
? Anonymous 2
Help please?!
NEW
Incorrect: If spring A is the heavy spring then the
transmitted pulse will travel slower than reflected pulse.
Correct: The only case in which there will be no reflected
pulse is when the springs are identical.
Incorrect: If spring A is the light spring then the
transmitted pulse will be inverted.
Incorrect: If spring A is the light spring then the
reflected pulse will not be inverted.
Correct: If spring A is the heavy spring then the reflected
pulse will not be inverted.
NEW
Thanks!!! Anonymous 4
Thank you so much!!!
Standing Waves
The length of the B string on a certain guitar is 59.0 . It vibrates at a frequency of 245.0 . What is the speed of the transverse waves on the string?
289 m/s
If the linear mass density of the guitar string is 1.60 what should the tension be when the string is in tune?
133.7 N
289 m/s
If the linear mass density of the guitar string is 1.60 what should the tension be when the string is in tune?
133.7 N
Discussion
NEW
Multiply your first answer times the density in kg/m
NEW
Square first ans. then mult. by density
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Re: Anonymous 3
How did you get the first answer?
NEW
part one
the velocity is :
2Lf ~ L is the length the the string in meters and f is
the frequency given.
Hope this helps!
NEW
Re: part one
thank you! and make sure you convert from cm to m
NEW
Units for part 2 Anonymous 7
Can I please get some help on the units for part 2?
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Re: Units for part 2 Anonymous 8
units im pretty sure are N.. can u explain how to find it?
NEW
Units Part 2
Wish I could help, but my answer was wrong
Does anyone have a simple explanation for part 2?
NEW
part 2 Anonymous 9
http://hyperphysics.phy-
astr.gsu.edu/hbase/Waves/string.html#c2
If you type in the values of frequency and velocity, length
and density into the calculator you will get the answer!
NEW
Concerning the second part...
...for the second part, take the answer from the first part
and square it. For me, this meant
(287 m/s)^2 = 82369 m^2/s^2
Take that number and multiply it by the given density,
except the density should be in kg/m rather than g/m. For me
this was
1.9 g/m -> 0.0019 kg/m
So..
(287 m/s)^2 * 0.0019 kg/m = 156 kg*m/s^2 or 156 N
Hope this proves helpful. :D
NEW
I got the answer right for part 1, based on V=2Lf, I'm just
trying to understand the logic behind it. I thought that
the speed of the wave was independent of the frequency, so
I'm not understanding why you would use the frequency to
find the answer to this problem. Any help would be
appreciated. Thanks!
A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, , is described by the following equation:
where x and y are in meters and the time is in seconds
What is the wavelength of the wave?
2.09 m
What is the velocity of the wave? (Define positive velocity along the positive x-axis.)
4.60×101 m/s
What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)
4.42 m/s
NEW
Anonymous
Hey, could someone help a bit with how to do the third part
of this problem?
NEW
Anonymous 2
How about the first part?
NEW
Re:
Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], lambda
= (2*pi)/k
NEW
Re:
Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], |v|
= omega*A
NEW
Re:
Hey Tacara! For the 3rd part max v is equal to omega times A.
A is the number right before "xsin" :)
NEW
second part Anonymous 6
How do we find the regular velocity for the second part?
NEW
Part 2 Anonymous 7
Can someone please explain how to do Part 2?
NEW
nvm..Part 2 Anonymous 7
For part 2, find k using lambda=6.28/k.
then, do w/k and that equals the velocity.
mine ended up being negative.
(w is from the eqn, Acos(kx − wt)
NEW
Part 2
I did V=W/k
K=lamda/2pi
W=#before T
Can anyone please tell me what I am doing wrong?
NEW
Re: Re: Anonymous 9 (
hey which value do we use for omega. is it the answer to
number 2?
I cant seem to figure out #3.
thanx
NEW
Omega is the number before t.
Can someone please post their answer for part 2, so I can
see if my answer is close.
NEW
Re:
for part 2, the answer is given by multiplying wavelength
(in meters) by the frequency (in s^-1). For my question,
this elicited
1.08 m * -24.8 s^-1
=-26.9 m/s
Because it is in the positive x-direction, the answer should
be negative. This gave
-(-26.9 m/s) or simply 26.9 m/s
Hope that helps.
NEW
Thanks
Thanks Daniel!!!
Multiply your first answer times the density in kg/m
NEW
Square first ans. then mult. by density
NEW
Re: Anonymous 3
How did you get the first answer?
NEW
part one
the velocity is :
2Lf ~ L is the length the the string in meters and f is
the frequency given.
Hope this helps!
NEW
Re: part one
thank you! and make sure you convert from cm to m
NEW
Units for part 2 Anonymous 7
Can I please get some help on the units for part 2?
NEW
Re: Units for part 2 Anonymous 8
units im pretty sure are N.. can u explain how to find it?
NEW
Units Part 2
Wish I could help, but my answer was wrong
Does anyone have a simple explanation for part 2?
NEW
part 2 Anonymous 9
http://hyperphysics.phy-
astr.gsu.edu/hbase/Waves/string.html#c2
If you type in the values of frequency and velocity, length
and density into the calculator you will get the answer!
NEW
Concerning the second part...
...for the second part, take the answer from the first part
and square it. For me, this meant
(287 m/s)^2 = 82369 m^2/s^2
Take that number and multiply it by the given density,
except the density should be in kg/m rather than g/m. For me
this was
1.9 g/m -> 0.0019 kg/m
So..
(287 m/s)^2 * 0.0019 kg/m = 156 kg*m/s^2 or 156 N
Hope this proves helpful. :D
NEW
I got the answer right for part 1, based on V=2Lf, I'm just
trying to understand the logic behind it. I thought that
the speed of the wave was independent of the frequency, so
I'm not understanding why you would use the frequency to
find the answer to this problem. Any help would be
appreciated. Thanks!
A transverse mechanical wave is traveling along a string lying along the x-axis. The displacement of the string as a function of position and time, , is described by the following equation:
where x and y are in meters and the time is in seconds
What is the wavelength of the wave?
2.09 m
What is the velocity of the wave? (Define positive velocity along the positive x-axis.)
4.60×101 m/s
What is the maximum speed in the y-direction of any piece of the string? (Give a positive answer for speed.)
4.42 m/s
NEW
Anonymous
Hey, could someone help a bit with how to do the third part
of this problem?
NEW
Anonymous 2
How about the first part?
NEW
Re:
Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], lambda
= (2*pi)/k
NEW
Re:
Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], |v|
= omega*A
NEW
Re:
Hey Tacara! For the 3rd part max v is equal to omega times A.
A is the number right before "xsin" :)
NEW
second part Anonymous 6
How do we find the regular velocity for the second part?
NEW
Part 2 Anonymous 7
Can someone please explain how to do Part 2?
NEW
nvm..Part 2 Anonymous 7
For part 2, find k using lambda=6.28/k.
then, do w/k and that equals the velocity.
mine ended up being negative.
(w is from the eqn, Acos(kx − wt)
NEW
Part 2
I did V=W/k
K=lamda/2pi
W=#before T
Can anyone please tell me what I am doing wrong?
NEW
Re: Re: Anonymous 9 (
hey which value do we use for omega. is it the answer to
number 2?
I cant seem to figure out #3.
thanx
NEW
Omega is the number before t.
Can someone please post their answer for part 2, so I can
see if my answer is close.
NEW
Re:
for part 2, the answer is given by multiplying wavelength
(in meters) by the frequency (in s^-1). For my question,
this elicited
1.08 m * -24.8 s^-1
=-26.9 m/s
Because it is in the positive x-direction, the answer should
be negative. This gave
-(-26.9 m/s) or simply 26.9 m/s
Hope that helps.
NEW
Thanks
Thanks Daniel!!!
Constructive and Destructive Interference
Two speakers, separated by a distance x = 7.12 are driven in phase by the same amplifier. A listener is located at point A, a distance L = 4.89 directly in front of one of the speakers. What is the lowest frequency for which there is a maximum signal at A due to constructive interference? The speed of sound can be taken as 343 .
91.5 Hz
What is the lowest frequency for which there is a minimum signal due to destructive interference?
45.8 Hz
91.5 Hz
What is the lowest frequency for which there is a minimum signal due to destructive interference?
45.8 Hz
Discussion
NEW
Anonymous 3
To get total interference, the number wavelengths along the 2.7 m path must
differ from the number of wavelengths along the angled path by an integral
number of wavelengths. The number of waves along the 2.7 m path is 2.7/L,
where L is the wavelength = V/f, where V = sound velocity and f = frequency.
The number of waves along the angle path is √[4.1^2 + 2.7^2] / L = 4.9092 /
L
For the lowest frequency, the number must differ by one wavelength, so
4.9092 / L - 2.7 / L = 1
L = 4.9092 - 2.7 = 2.2092 m
For the speed of sound = 331.3 m/s, this gives f = v/L = 150 Hz
NEW
Hope this helps Anonymous 6
Part 1
I drew out the diagram that they gave us and it made a
triangle between the three points (listener, speaker 1, and
speaker 2) I then figured out the hypotenuse (H).
H - L(given) = L (constructive intereference)
V is given to be 343 m/s
f= V/ L (constructive intereference) and that will give you
the frequency
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Re: Hope this helps Anonymous 8
How did you do part 2? Thanks!
NEW
Re: Re: Hope this helps Anonymous 9
double the L
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Re: Re: Re: Hope this helps Anonymous 10
which L am i doubling... i have tried the given and constructive one but neither
worked
NEW
Re: Re: Re: Re: Hope this helps Anonymous 11
Divide your first answer by 2.
Anonymous 3
To get total interference, the number wavelengths along the 2.7 m path must
differ from the number of wavelengths along the angled path by an integral
number of wavelengths. The number of waves along the 2.7 m path is 2.7/L,
where L is the wavelength = V/f, where V = sound velocity and f = frequency.
The number of waves along the angle path is √[4.1^2 + 2.7^2] / L = 4.9092 /
L
For the lowest frequency, the number must differ by one wavelength, so
4.9092 / L - 2.7 / L = 1
L = 4.9092 - 2.7 = 2.2092 m
For the speed of sound = 331.3 m/s, this gives f = v/L = 150 Hz
NEW
Hope this helps Anonymous 6
Part 1
I drew out the diagram that they gave us and it made a
triangle between the three points (listener, speaker 1, and
speaker 2) I then figured out the hypotenuse (H).
H - L(given) = L (constructive intereference)
V is given to be 343 m/s
f= V/ L (constructive intereference) and that will give you
the frequency
NEW
Re: Hope this helps Anonymous 8
How did you do part 2? Thanks!
NEW
Re: Re: Hope this helps Anonymous 9
double the L
NEW
Re: Re: Re: Hope this helps Anonymous 10
which L am i doubling... i have tried the given and constructive one but neither
worked
NEW
Re: Re: Re: Re: Hope this helps Anonymous 11
Divide your first answer by 2.