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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 cartridge 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 stylus and cartridge.

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  • \$\begingroup\$ Can't you test that yourself? Run the 33 record at 33 and 45 rpm. Record both with your PC. Then time-stretch the fast one by 33/45. Calculate the difference of the signals to see what changed. \$\endgroup\$ – Thomas Weller Apr 6 at 19:54
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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.

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    \$\begingroup\$ +1 Bingo! This is the best answer for someone "...very interested in vinyl records and analog music." \$\endgroup\$ – uhoh Apr 6 at 4:32
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    \$\begingroup\$ Far and away the best answer. (As @uhoh says!) \$\endgroup\$ – Fattie Apr 6 at 15:35
  • \$\begingroup\$ Good answer, however the pitch shift itself is actually going to be by far the most notable consequence of the speed change. Human ears are very sensitive to pitch, but compensate pretty well for frequency response. Especially, since the RIAA curve is a relatively straight, monotonous descend, shifting it mostly just changes the total level and only a little bit the frequency response. \$\endgroup\$ – leftaroundabout Apr 7 at 10:49
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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.

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  • \$\begingroup\$ What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies. \$\endgroup\$ – Henning Makholm Apr 6 at 13:16
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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.

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  • \$\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 Apr 5 at 22:20
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    \$\begingroup\$ It's linear -- double the speed means double the frequency. That's what compressing the time axis means. \$\endgroup\$ – Dave Tweed Apr 5 at 22:21
  • \$\begingroup\$ Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown. \$\endgroup\$ – Henning Makholm Apr 6 at 13:21
  • \$\begingroup\$ @HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that. \$\endgroup\$ – Dave Tweed Apr 6 at 13:53
  • \$\begingroup\$ @DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is). \$\endgroup\$ – Henning Makholm Apr 6 at 14:12
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To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, 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.

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    \$\begingroup\$ I was surprised when I first learned how records work. It's so analog that it's amazing it works at all. \$\endgroup\$ – DKNguyen Apr 5 at 22:27
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    \$\begingroup\$ @Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past. \$\endgroup\$ – Graham Apr 6 at 7:16
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I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.

Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.

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  • \$\begingroup\$ Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument \$\endgroup\$ – DJohnM Apr 6 at 6:18
  • \$\begingroup\$ @DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves. \$\endgroup\$ – Graham Apr 6 at 7:03
  • \$\begingroup\$ DJs use variable-speed players too, to match the tempo (beat) between records. Players like the Technics SL 1200 were/are popular among that crowd. \$\endgroup\$ – Hobbes Apr 6 at 19:47
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The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.

Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil. The high resistance causes a high random thermal electron noise floor.

The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.

But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less, at that stated needle velocity, at 1KHz.

[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.

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