Chapter 10.3 Quiz
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Chapter 10.3 Quiz
I know it's early, but just to get a headstart on the weekend's work!
1.) The traveling wave velocity on a 0.4335 mlong string is 171.7 m/s. What is the frequency of the 5 th harmonic (see sheet 19). Indicate with a positive (negative) sign whether the frequency you calculated is (is not) an integer multiple of the fundamental frequency ?
2.) The standing sound wave in a pipe with one end open has a fundamental frequency of 482.9 Hz. What is the length of the pipe (see sheet 26,use 343 m/s for the speed of sound) ? Indicate with a negative (positive) sign whether all (only odd) integer multiples of the fundamental frequency are allowed.
3.) A piano tuner hears a beat frequency of 7.074 Hz when exciting a tuning fork with a 526.4 Hz fundamental frequency close to a piano string. (Assume that the fundamental frequency of the tuning fork is the dominant frequency exciting the piano string.) What is the larger of the two possible string frequencies (see sheet 30,31') ? Indicate with a negative (positive) sign whether you can (can not) calculate the other possible string frequency from the data given.
4.)A string under tension is 1.4561 m long and weighs 1.757 grams. The frequency of the second harmonic standing wave on the string is 274.6 Hz. What is the tension of the string (see sheet 20,22) (Watch out! It is the mass per unit string length and not the total string mass that enters your calculation) ? Indicate with a negative (positive) sign whether the tension is higher (lower) if the frequency above is the fundamental frequency.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Question 1:
Answer is POSITIVE because if you apply the fundamental frequency equation (f=v/2L), and then apply the one for f5 (f_5=5*(v/2L)) you find out the f5 answer is just the answer from the fundamental frequency multiplied by 5.
f_5 = 5* v/2L
f_5 = 5* (171.7/2*.4335)
f_5 = 990.196
1.) The traveling wave velocity on a 0.4335 mlong string is 171.7 m/s. What is the frequency of the 5 th harmonic (see sheet 19). Indicate with a positive (negative) sign whether the frequency you calculated is (is not) an integer multiple of the fundamental frequency ?
2.) The standing sound wave in a pipe with one end open has a fundamental frequency of 482.9 Hz. What is the length of the pipe (see sheet 26,use 343 m/s for the speed of sound) ? Indicate with a negative (positive) sign whether all (only odd) integer multiples of the fundamental frequency are allowed.
3.) A piano tuner hears a beat frequency of 7.074 Hz when exciting a tuning fork with a 526.4 Hz fundamental frequency close to a piano string. (Assume that the fundamental frequency of the tuning fork is the dominant frequency exciting the piano string.) What is the larger of the two possible string frequencies (see sheet 30,31') ? Indicate with a negative (positive) sign whether you can (can not) calculate the other possible string frequency from the data given.
4.)A string under tension is 1.4561 m long and weighs 1.757 grams. The frequency of the second harmonic standing wave on the string is 274.6 Hz. What is the tension of the string (see sheet 20,22) (Watch out! It is the mass per unit string length and not the total string mass that enters your calculation) ? Indicate with a negative (positive) sign whether the tension is higher (lower) if the frequency above is the fundamental frequency.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Question 1:
Answer is POSITIVE because if you apply the fundamental frequency equation (f=v/2L), and then apply the one for f5 (f_5=5*(v/2L)) you find out the f5 answer is just the answer from the fundamental frequency multiplied by 5.
f_5 = 5* v/2L
f_5 = 5* (171.7/2*.4335)
f_5 = 990.196
Kathleen Guest
Question 2
NOTE: For Question 1, I didn't realize this but the harmonic value changes each quiz. So although I multiplied by 5 when it asked for frequency of the 5th harmonic, they may also ask for 6th, 7th, etc. In which case you have to apply the same v/2L formula and then multiply by either the 6 or 7  whatever is asked of you.
Question 2:
It asks for fundamental frequency, so its f = v/4L
4L = v/f
v/f = x
L = x*4
Answer is POSITIVE because this can only be used for odd numbers (as stated on notes from cd on sheet 26)
Question 2:
It asks for fundamental frequency, so its f = v/4L
4L = v/f
v/f = x
L = x*4
Answer is POSITIVE because this can only be used for odd numbers (as stated on notes from cd on sheet 26)
Kathleen Guest
Re: Chapter 10.3 Quiz
Hey Kathleen, for Question 2 instead of multiplying x by 4 you should be dividing x by 4
Question 3
Answer is Negative since you can find the other frequency by switching signs.
Very simple, it is asking for the larger frequency.
Frequency #2 + Frequency #1
Question 4
Answer is Negative
First convert your grams into kilograms.
Now you have to find both v and mu.
mu = m/L
v = L * f
Now to get the period, T
T = mu*v^{2}
Tada!
Question 3
Answer is Negative since you can find the other frequency by switching signs.
Very simple, it is asking for the larger frequency.
Frequency #2 + Frequency #1
Question 4
Answer is Negative
First convert your grams into kilograms.
Now you have to find both v and mu.
mu = m/L
v = L * f
Now to get the period, T
T = mu*v^{2}
Tada!
Guest01 Posts : 133
Join date : 20080919
Chapter 10.3 Quiz
So the information for number two is all correct except for x*4. If you solve for L you need to divide by 4 so...
f=v/4L
4L=v/f
L=v/(f*4)
OR
f=v/4L
4L=v/f
v/f=x
L=x/4
f=v/4L
4L=v/f
L=v/(f*4)
OR
f=v/4L
4L=v/f
v/f=x
L=x/4
Guest Guest
Re: Chapter 10.3 Quiz
i dont understand how to find "mu". i know that "L" is the string under tension and i dont understand "m" in the eequation mu=m/L.
question Guest
question 2
Yeah I'm sorry, I typed it wrong though I did get it right on maple ta!
Thank you for the correction though, so now at least others will understand it better.
Thank you for the correction though, so now at least others will understand it better.
Kathleen Guest
Re: Chapter 10.3 Quiz
I was lazy, "mu" is the Greek letter mu (μ). m is the mass in kilograms and L is just the length of the string. The equation should really look like T = μ*v^{2}. You can also just follow page 22 on the CD to the tee.
Guest01 Posts : 133
Join date : 20080919
Re: Chapter 10.3 Quiz
Make sure for Question 4 it says "The frequency of the second harmonic standing wave on the string is xxx.x Hz". If it says this then using the formula f_{n} = n(v/2L) becomes f_{2} = 2(v/2L) and the 2's cancel out leaving f = v/L. Rearrange to solve for v gives us v = L*f
Then you can solve for tension. Make sure you're in Kilograms. If you are still having problems can you post your work?
Then you can solve for tension. Make sure you're in Kilograms. If you are still having problems can you post your work?
Guest01 Posts : 133
Join date : 20080919
Re: Chapter 10.3 Quiz
For question 4, i am confused about the +/. Comparing the fundamental frequency to the second harmonic motion, do you not have a SMALLer velocity resulting into a smaller tension so its Lower? Can you show me how you get a higher velocity? thanks.
Jman Guest
Question 1 help
Okay this is your help for question 1
f = v/2*L
where v = the traveling wave velocity (some number over 100)
L = string length (a decimal .xxxx m)
so do v divided by 2*L = Z
and then in your question it will say something about the harmonic motion, whether it is the 5th, 6th, 7th, etc.
Depending on what number it says, that is the number you multiply by Z
if that doesn't help, post your question
f = v/2*L
where v = the traveling wave velocity (some number over 100)
L = string length (a decimal .xxxx m)
so do v divided by 2*L = Z
and then in your question it will say something about the harmonic motion, whether it is the 5th, 6th, 7th, etc.
Depending on what number it says, that is the number you multiply by Z
if that doesn't help, post your question
Kathleen Guest
Re: Chapter 10.3 Quiz
Jman I think you are misread the question. It says, "Indicate with a negative (positive) sign whether the tension is higher (lower) if the frequency above is the fundamental frequency."
meaning, if you are given the 2nd harmonic in the original question, pretend it is now the fundamental frequency (meaning, "n" decreased)
meaning, if you are given the 2nd harmonic in the original question, pretend it is now the fundamental frequency (meaning, "n" decreased)
an0n Guest
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