## 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.

What is the wavelength of the traveling wave?

What is the period of this wave?

**-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

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**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

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**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.

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**Re: Re: Speed**

Thanks

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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?

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**Where to start?**

*Anonymous 4*

Can someone please help me to understand where I start on

this? Thank you so much for your help!

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**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!

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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

## 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**## Discussion

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**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

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**% 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???

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**Re: not working?**

*Anonymous 6*

maybe you forgot the negative sign in your answer?

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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!

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**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.## Discussion

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**Help**

Not sure about this one. Any pointers?

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**?**

*Anonymous 2*

Help please?!

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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?

If the linear mass density of the guitar string is 1.60 what should the tension be when the string is in tune?

**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

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Multiply your first answer times the density in kg/m

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Square first ans. then mult. by density

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**Re:**

*Anonymous 3*

How did you get the first answer?

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**part one**

the velocity is :

2Lf ~ L is the length the the string in meters and f is

the frequency given.

Hope this helps!

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**Re: part one**

thank you! and make sure you convert from cm to m

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**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?

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**Units Part 2**

Wish I could help, but my answer was wrong

Does anyone have a simple explanation for part 2?

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**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**

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*Anonymous*

Hey, could someone help a bit with how to do the third part

of this problem?

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*Anonymous 2*

How about the first part?

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**Re:**

Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], lambda

= (2*pi)/k

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**Re:**

Given the equation [ y(x, t) = A*sin(k*x − omega*t) ], |v|

= omega*A

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**Re:**

Hey Tacara! For the 3rd part max v is equal to omega times A.

A is the number right before "xsin" :)

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**second part**

*Anonymous 6*

How do we find the regular velocity for the second part?

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**Part 2**

*Anonymous 7*

Can someone please explain how to do Part 2?

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**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)

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**Part 2**

I did V=W/k

K=lamda/2pi

W=#before T

Can anyone please tell me what I am doing wrong?

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**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 .

What is the lowest frequency for which there is a minimum signal due to destructive interference?

**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!

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**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

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**Re: Re: Re: Re: Hope this helps**

*Anonymous 11*

Divide your first answer by 2.