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What's the output of a record needle playing an out-of-speed record

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2

$begingroup$

I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.

brushless-dc-motor

share|improve this question

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

$endgroup$

add a comment | 

2

$begingroup$

I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.

brushless-dc-motor

share|improve this question

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

$endgroup$

add a comment | 

$begingroup$

I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.

brushless-dc-motor

share|improve this question

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

$endgroup$

share|improve this question

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

share|improve this question

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

edited 1 hour ago

Dave Tweed♦

123k9152266

edited 1 hour ago

Dave Tweed♦

123k9152266

asked 1 hour ago

Gabriel SantosGabriel Santos

213

asked 1 hour ago

Gabriel SantosGabriel Santos

213

4 Answers
4

active

oldest

votes

5

$begingroup$

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 33¹_/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.

The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

$endgroup$

add a comment | 

3

$begingroup$

To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).

Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.

If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
  $endgroup$
  – Toor
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.

A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
  $endgroup$
  – Gabriel Santos
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
  $endgroup$
  – Dave Tweed♦
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.

In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.

share|improve this answer

answered 21 mins ago

user217611

$endgroup$

add a comment | 

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5

$begingroup$

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 33¹_/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.

The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

$endgroup$

add a comment | 

5

$begingroup$

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 33¹_/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.

The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

$endgroup$

add a comment | 

$begingroup$

Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?

The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 33¹_/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.

The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

$endgroup$

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

edited 1 hour ago

K H

2,360215

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TransistorTransistor

88.2k785189

answered 1 hour ago

TransistorTransistor

88.2k785189

3

$begingroup$

To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).

Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.

If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
  $endgroup$
  – Toor
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).

Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.

If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
  $endgroup$
  – Toor
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).

Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.

If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

$endgroup$

share|improve this answer

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

edited 1 hour ago

K H

2,360215

edited 1 hour ago

K H

2,360215

answered 1 hour ago

TimWescottTimWescott

6,5991416

answered 1 hour ago

TimWescottTimWescott

6,5991416

2

$begingroup$

Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.

A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
  $endgroup$
  – Gabriel Santos
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
  $endgroup$
  – Dave Tweed♦
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.

A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
  $endgroup$
  – Gabriel Santos
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
  $endgroup$
  – Dave Tweed♦
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.

A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

$endgroup$

share|improve this answer

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266

answered 1 hour ago

Dave Tweed♦Dave Tweed

123k9152266