# What does it mean to make a sound twice as loud? [duplicate]

Giancoli Textbook question:

To make a given sound twice as loud, how should a musician change the intensity of the sound?

The given answer is: "Increase the intensity by a factor of 10."

I don't get it, doesn't increasing it by a factor of 10 contribute +10 decibels and is only correct if the sound is 10 decibels from the start? I'm thinking that it might not be referring to the sound level in dB, but if it does not refer to the sound level, what does it refer to? Thanks for the help.

## marked as duplicate by Rococo, Jon Custer, Kyle Kanos, stafusa, ZeroTheHeroOct 17 '18 at 21:51

In psychoacoustic experiments, when asked to turn up the volume twice as high, most people increase the sound level with about 10 dB which is 10 times the intensity.

It is not really a physics thing. It is perception. And natural language.

Edit: The textbook mentions it. In section 12-2 Giancoli writes: "To produce a sound that sounds about twice as loud requires a sound wave that has about 10 times the intensity."

He goes on to say that "four times as loud" is $$100$$ times the intensity. I would be a bit skeptical about that.

• @Pieter, you have clearly identified the distinction required to answer the question posed by the OP. Intensity is the amount of power per area (watts/$m^2$). Intensity in decibels is 10 x log of the ratio of a sound to a standard intensity. Loudness is a perceptual/subjective experience related to the saturation of receptor and consciousness. hyperphysics.phy-astr.gsu.edu/hbase/Sound/intens.html#c1 – Thomas Lee Abshier ND Oct 9 '18 at 14:56
• Pieter is right about this one. Loudness is about perception, and that is what he said. He never said intensity measurements and the decibels scale were perception. Aaron Stevens, I think you did not understand what Pieter wrote. – Hugo V Oct 9 '18 at 14:56
• Thanks. But I totally agree with the answer by @HugoV, I upvoted it. I think my short answer emphasizes Giancoli's point a bit better. And the OP's difficulty. – Pieter Oct 9 '18 at 17:02
• It sounds to me like there must be something in the textbook explaining this, otherwise how would a student be expected to know this answer? So I think what would make this a really great answer is to find that passage in the book and add a quote of it. Of course, I understand that it's probably not practical to do that if you don't actually have a copy of the book. Without that, this seems like about as good of an answer as it's possible to give. – David Z Oct 9 '18 at 18:58
• @aquirdturtle A logarithmic dependence would be Fechner's law, but that does not tell us whether "twice as loud" is 10 dB or 7 dB or 15 dB. There is also en.wikipedia.org/wiki/Stevens%27s_power_law – Pieter Oct 10 '18 at 7:58

I think you are confusing the relationship between loudness and sound intensity, with the relationship between Sound Intensity Level and sound intensity relative to a reference value, that is expressed by the equation: $$L_I=10 \log_{10}\left(\frac I{I_0}\right) \textrm{dB}$$ What you are talking about in your post is the relationship between loudness, which is the strength of ear's perception of sound, and the physical quantity called the intensity of sound. They are related by the "rule of thumb" that you just described. You can find this information at: http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/loud.html

• uhm, i just read the link, why do 10 times the intensity result in twice the sound level in db?, since b= 10log(10 intensity/ Intensity not) will just contribute + 10 decibels. – SuperMage1 Oct 9 '18 at 14:31
• It does not. Twice as loud does not mean twice the sound level. Loudness is the perception of the sound by the ears, it depends on how the ears work, while sound level is as physical quantity that depends only on the sound wave itself. The fact that you hear twice as loud something that has 10 times the intensity is an empirical fact, and is only a "rule of thumb", an approximation, it depends on the person hearing it and on the frequency of the wave in question. – Hugo V Oct 9 '18 at 15:21
• I think that "loudness" may have a technical interpretation, with dB units. But "twice as loud" is just non-scientific language, like "twice as high". – Pieter Oct 9 '18 at 17:18

As others noted, loudness is a psycho-physical sensation perceived by the human auditory perception. It is not the same as sound pressure level, in which doubling the intensity increases the spl by 3db (not 10db), or voltage by 6dB.

See this plot:

Source

Therefore:

Double or twice the loudness [intensity] = factor 2 means about 10 dB more sensed loudness level (psycho acoustic)

Double or twice the voltage [intensity]= factor 2 means 6 dB more measured voltage level (sound pressure level)

Double or twice the power [intensity]= factor 2 means 3 dB more calculated power level (sound intensity level)

EDIT:

Below are the equations for Loudness with respect to intensity ratio (as well as voltage and power) As you can see, the equation governing loudness is log2 as opposed to the traditional log10. Though it can be represented in log10, the multiplier becomes 33.22 (instead of the traditional 10 or 20 used for power and voltage, respectively)

• I doubt that an instruction to set the volume eight times as high would mean much to people. I suspect that very few would get close to 30 dB. – Pieter Oct 9 '18 at 22:00
• @Pieter That took me a while to get. Yes, 8 =2^3 -> 3*10dB = 30dB. So yes, if someone told me to set the volume 8 times as high, I would have to increase the volume (loudness) by 30dB. That does beg the question what the numbers on a volume dial represent. 10dB intervals wouldn't make too much sense, but linear volume (non-log scale) doesn't make much sense either. I would suspect they are voltage level, which is not intuitive how it relates to volume. – dberm22 Oct 9 '18 at 22:18
• @dberm22 Try working the volume dial on a few of your devices, especially where multiple levels are applied in a succession. You'll find precious few that have volume controls that make sense beyond "go right for more loud". – Luaan Oct 10 '18 at 5:37
• Even on a log-log chart, perceived loudness won't be a straight line. In the presence of ambient sound, even a 3dB change in signal level can make a large difference in perceived loudness. – supercat Oct 10 '18 at 16:31

Giancoli's question and answer are simply nonsense.

If "twice as loud" is interpreted in physical terms, then it means twice the intensity, in units of watts per square meter, and this is then a change of about 3 dB, not 10 dB.

If, on the other hand, "twice as loud" is supposed to be interpreted in psychoacoustic terms, then it's still not really very meaningful, or meaningful only in the sense that if you poll people and ask them to judge what is "twice as loud," their responses, when measured physically, may cluster around some number of dB -- but there is certainly no way to predict this based on having read a freshman physics textbook. This is like asking someone whether something is "twice as sour," or if a puppy is "twice as cute," or if a certain shade of red is "twice as saturated," or if Trump is "twice as evil" as George W. Bush.

Having had the misfortune to have taught from Giancoli, I'm not surprised that there is crap this bad in that book, but wow...this is shockingly bad.

• I looked it up. In section 12-2 it says: "To produce a sound that sounds about twice as loud requires a sound wave that has about 10 times the intensity," It also says that "four times as loud" is 100 times the intensity. I would be a bit skeptical about that. – Pieter Oct 10 '18 at 10:00
• @Pieter: Then Giancoli is inculcating the "puppy that's twice as cute" misconception in his readers. – Ben Crowell Oct 10 '18 at 14:08
• Psychophysical experiments may give surprisingly consistent answers to what people experience as "twice as hot", "twice as red". People can rank dogs for cuteness. (As for politicians: Bush started wars, Trump so far has not.) – Pieter Oct 10 '18 at 15:29

As already mentioned, decibels are on a logarithmic scale: $$L=10\log_{10}\left(\frac{I}{I_0}\right)\ \rm{dB}$$

If we do what the answer says: increase $$I$$ by a factor of $$10$$, what happens? $$L'=10\log_{10}\left(\frac{10I}{I_0}\right)\ \rm{dB}=10\log_{10}\left(\frac{I}{I_0}\right)\ \rm{dB}+10\ \rm{dB}$$

So we see that we actually add $$10\ \rm{dB}$$ to the decibel rating. Whether or not this is "twice as 'loud'" is somewhat subjective, as others have pointed out as well.

If we wanted to actually double the decibels we would have $$2L=10\times2\log_{10}\left(\frac{I}{I_0}\right)\ \rm{dB}=10\log_{10}\left(\frac{I^2/I_0}{I_0}\right)\ \rm{dB}$$

So you would want your intensity to become $$I'=\frac{I^2}{I_0}$$ (i.e., increase by a factor of $$\frac{I}{I_0}$$)

Which does in fact depend on the initial intensity $$I$$.

So really either the question could have been written better if this information is not in the book (since it seems like it is looking for a single quantitative correct answer, when there isn't one), or the solution could have gone into what assumptions were being made to arrive at the answer of multiplying the intensity by $$10$$.

• That is certainly not true. And it is what you were getting wrong in the comments. $L' \neq 2L$. That would mean $2L=L+10$, hence, $L=10$, so it would only be true for a certain especific sound level. The whole point is that loudness is not the same as sound level. 10 times the intensity does make 2 times the sound level. – Hugo V Oct 9 '18 at 15:25
• @HugoV sorry I was in a rush. I didn't mean to put the $2L$ in. I was going to approach from the OPs view of actually wanting to double the decibels and seeing what this actually meant for the intensity, but then I decided to switch approaches and forgot to remove the $2L$ – Aaron Stevens Oct 9 '18 at 15:50
• It is not about doubling decibels. "Twice as loud" does not have units. Just like putting the radio twice as high does not mean to multiply its distance from the floor by a factor of two. – Pieter Oct 9 '18 at 16:56
• The interesting thing is that it is not purely subjective. What "twice as loud" is can be determined by psychoacoustic experiments. The results are surprisingly consistent. – Pieter Oct 9 '18 at 17:34
• But there is nothing wrong with Giancoli's question. As @David Z says, the book probably mentions somewhere that an increase by about 10 dB in sound level is generally perceived as twice as loud. – Pieter Oct 9 '18 at 19:50

Don't try to establish repeatable settings based on the human hearing response 'trends' documented in psychoacoustics. Psychoacoustics is variable as each person's hearing response is widely variant.

If you play two pure notes 500 Hertz and 600 Hertz at the same amplitude at a comfortable listening level, then adjust the amplitude of one note down, you will reach a point were the lower amplitude sound level note will not be discerned by natural hearing. Yet most likely if you turned off the higher amplitude note, you would then hear the lower amplitude note just fine. Now set the two notes closer in frequency, 500 Hertz and 525 Hertz. Begin adjusting the amplitude of one note downward and you will discover you can no longer hear the lower amplitude note when it is still at a much higher amplitude when compared to the lower amplitude note more widely separated in frequency. This phenomena is often distorted in individuals with hearing loss or other auditory differences in their hearing. If you adjust the starting amplitude up or down when measuring amplitude differential using the loss of perception in the one note as the metric you will find the amplitude differential between the two notes will vary dramatically.

Psychoacoustics is a good tool to understand there why there is a differential in hearing response but to compare sound levels you still need to rely on physically measured properties and not the widely variant individual perception when quantifying properties.