Two basic physics 11 wave questions

[Kingsley]

QUESTION 1

Homework Statement

A guitar string is 0.70m long and is tuned to play an E above middle C (330 Hz). How far from the end of this string must a finger be placed to play A above middle C (440Hz)

Homework Equations

\lambda = 2L/n
\lambda = v/f
v = \lambda x f

The Attempt at a Solution

\lambda = 2L/n
\lambda = 2(0.70m)/1
\lambda = 1.4m

v = \lambda x f
v = (1.4m)(330Hz)
v = 462m/s

\lambda = v/f
\lambda = 462m/s/440Hz
\lambda = 1.05

\lambda = 2L/n
\lambda = L/2
\lambda = 1.05m/2
\lambda = 0.525m

that is the right math i don't know what to add to find how far to place your finger though, help/explanation would be appreciated.






QUESTION 2

Homework Statement

If you were to build a pipe organ with closed tube pipes spanning the range of human hearing (20 Hz to 20 kHz), what would be the range of pipes required?

Homework Equations

\lambda = 2L/n
\lambda = v/f
v = \lambda x f
\lambda = 2L/n <<< SIDE NOTE: what's the difference between
\lambda = 4L/2n - 1 these two equations?

The Attempt at a Solution

v/f = \lambda
\lambda = 340m/s/20Hz
\lambda = 17m

\lambda = 340m/s/20000Hz
\lambda = 0.017m

i have no idea what to do after this any help would be appreicated. thanks in advance.

 

[berkeman]

QUESTION 1

Homework Statement

A guitar string is 0.70m long and is tuned to play an E above middle C (330 Hz). How far from the end of this string must a finger be placed to play A above middle C (440Hz)

Homework Equations

\lambda = 2L/n
\lambda = v/f
v = \lambda x f

The Attempt at a Solution

\lambda = 2L/n
\lambda = 2(0.70m)/1
\lambda = 1.4m

v = \lambda x f
v = (1.4m)(330Hz)
v = 462m/s

\lambda = v/f
\lambda = 462m/s/440Hz
\lambda = 1.05

\lambda = 2L/n
\lambda = L/2
\lambda = 1.05m/2
\lambda = 0.525m

that is the right math i don't know what to add to find how far to place your finger though, help/explanation would be appreciated.






QUESTION 2

Homework Statement

If you were to build a pipe organ with closed tube pipes spanning the range of human hearing (20 Hz to 20 kHz), what would be the range of pipes required?

Homework Equations

\lambda = 2L/n
\lambda = v/f
v = \lambda x f
\lambda = 2L/n <<< SIDE NOTE: what's the difference between
\lambda = 4L/2n - 1 these two equations?

The Attempt at a Solution

v/f = \lambda
\lambda = 340m/s/20Hz
\lambda = 17m

\lambda = 340m/s/20000Hz
\lambda = 0.017m

i have no idea what to do after this any help would be appreicated. thanks in advance.

Welcome to the PF. On the first question, I don't think I'd bother calculating the velocity of propagation on the string. When the string is plucked, how many wavelengths are there between the two fixed ends? And when you place your finger on a fret to shorten the string to get the higher note, how many wavelengths are there on the vibrating string between the low end and your finger?

On the second question, are the pipes on an organ a full wavelength long? I honestly don't know, but you could easily google or wikipedia search to figure that out. The rest of what you are doing there looks like a good approach.

 

[Kingsley]

Welcome to the PF. On the first question, I don't think I'd bother calculating the velocity of propagation on the string. When the string is plucked, how many wavelengths are there between the two fixed ends? And when you place your finger on a fret to shorten the string to get the higher note, how many wavelengths are there on the vibrating string between the low end and your finger?

On the second question, are the pipes on an organ a full wavelength long? I honestly don't know, but you could easily google or wikipedia search to figure that out. The rest of what you are doing there looks like a good approach.

i don't know for the first one would you subtract 0.7m - 0.525m?

and for the second one you would use \lambda = 4L/2n-1 would the answer be 4.25m to 0.00425m?

i had both the right answers but i eraced them and i can't tell what the final answer would be for both them