How Does Temperature Affect the Pitch of a Violin Played Outdoors?

[BizzPhizz]

Homework Statement

Question 1: A violin is tuned indoors to play the proper frequencies. A musician then uses the violin to play outdoors where the temperature is considerably colder. Wil the violin play too sharp (high) or too flat (low)? Ignore temperature effects on length and density

The Attempt at a Solution

The frequency and wavelength are proportional to velocity in air. This means that velocity if speed in air decreases, then both wavelength and frequency would decrease. Since there is a temperature drop outside the speed of sound in air would decrease and thus ultimately cause wavelength and frequency of a wave traveling in air to decrease.

Since frequency is directly related to pitch, you would be hearing the violin play a flatter sound than if it was indoors because the frequency at which it plays outdoors would decrease.

Homework Statement

Question 2: An organ pipe plays a certain fundamental frequency when the pipe is open at both ends. If one end of the pipe is covered, what happens to the pitch of the note emitted?

The Attempt at a Solution

The condition for a fundamental frequency to be played in an open air column is that half the wavelength is equal to the length of the pipe. When you cover one end the pipe still stays the same length but the conditions change. You would not here any resonance because the node forms at both the open and closed ends of the air column since the air column is twice the size of the closed ended air columns fundamental frequency condition (l=/4)

 

[haruspex]

This means that velocity if speed in air decreases, then both wavelength and frequency would decrease.

Why both? When a note is played on a violin string, it plays with a certain wavelength and a certain frequency. You're told to ignore the temperature effect on the string, so these don't change. Are these the wavelength and frequency of the note in the air? If not, how are those determined by the wavelength and frequency at the string?

You would not here any resonance because the node forms at both the open and closed ends of the air column

Not sure what you mean by the bit about resonance, but I disagree with the second part. Where are the nodes and antinodes of the fundamental frequency in (a) a doubly open pipe (b) a singly open pipe (c) a closed pipe?

 

[BizzPhizz]

Why both? When a note is played on a violin string, it plays with a certain wavelength and a certain frequency. You're told to ignore the temperature effect on the string, so these don't change. Are these the wavelength and frequency of the note in the air? If not, how are those determined by the wavelength and frequency at the string?
I understand that nothing in the string changes, but of course you can't hear the note being played without the vibrations in the string causing air to vibrate at the same frequency. And if the temperature of the air is now different, wouldn't that affect the frequency you hear?

Is it because particles (under colder conditions) vibrate slower so they don't transfer energy as efficient? Or is it because of damping? Please Help me :/

Instead of answering my question with another question, can you just give me an answer? I mean, all I am asking for is if the pitch at an antinode is higher or lower than the pitch at a node. For the second Q

Thanks,
BizzPhizz

 

[haruspex]

I understand that nothing in the string changes, but of course you can't hear the note being played without the vibrations in the string causing air to vibrate at the same frequency. And if the temperature of the air is now different, wouldn't that affect the frequency you hear?

What is the air 'aware of'? The length of the wave in the string, or the frequency at which it vibrates, or both?

Instead of answering my question with another question, can you just give me an answer?

The preferred approach in these forums (and a principle to which I subscribe) is to provide just enough hints and leads to get you to reason it out for yourself. You'll get more out of it that way.

all I am asking for is if the pitch at an antinode is higher or lower than the pitch at a node.

From that I conclude you don't know what nodes and antinodes are.
In a standing wave, a node is a point at which maximum amplitude occurs, while an antinode is a point where the amplitude is zero. A string in its fundamental will have an antinode at each end and a node in the middle, so the whole string represents one half wavelength. Can you figure out the corresponding arrangements for pipes, according to the number of open ends?

 

[BizzPhizz]

I know what nodes and antinodes are, but I am kind of representing the questions not only I have, but my class as well. They ask me to post this stuff to help them out as well.

I know that under the condition a standing wave is locked between two points that Ln=n(λ/2) where n represents the harmonic.

In a closed ended air column, the length of air column required for resonance is Ln=(2n-1)λ/4 where the wave length can only be odd integers of λ/4

Ex: L=λ/4, <- First harmonic 3λ/4, <- Third Harmonic 5λ/4 <- Fifth Harmonic

Two open ends is the same corresponding as 2 closed ends.

 

[haruspex]

I know what nodes and antinodes are, but I am kind of representing the questions not only I have, but my class as well. They ask me to post this stuff to help them out as well.

Are you saying you posted a question that you knew made no sense, just because someone asked you to? You couldn't have set them straight yourself?
If you know what nodes and antinodes are, why did you not correctly answer my question: Where are the nodes and antinodes of the fundamental frequency in (a) a doubly open pipe (b) a singly open pipe (c) a closed pipe?
You need to be able to answer that to get the right answer to Q2

I know that under the condition a standing wave is locked between two points that Ln=n(λ/2) where n represents the harmonic.

"Locked"? If by that you mean the ends are both antinodes, yes.

In a closed ended air column, the length of air column required for resonance is Ln=(2n-1)λ/4 where the wave length can only be odd integers of λ/4

That's true for one end open, one closed.

Two open ends is the same corresponding as 2 closed ends.

In terms of harmonics, yes, but the node/antinode arrangements that generate the harmonics are different.