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What does it mean to make a sound twice as loud?
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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.
homework-and-exercises waves acoustics intensity
edited 21 mins ago
David Z♦
62.3k23135251
asked 5 hours ago
SuperMage1
1092
add a comment |Â
up vote
1
down vote
favorite
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.
homework-and-exercises waves acoustics intensity
edited 21 mins ago
David Z♦
62.3k23135251
asked 5 hours ago
SuperMage1
1092
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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.
homework-and-exercises waves acoustics intensity
edited 21 mins ago
David Z♦
62.3k23135251
asked 5 hours ago
SuperMage1
1092
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.
homework-and-exercises waves acoustics intensity
homework-and-exercises waves acoustics intensity
edited 21 mins ago
David Z♦
62.3k23135251
asked 5 hours ago
SuperMage1
1092
edited 21 mins ago
David Z♦
62.3k23135251
asked 5 hours ago
SuperMage1
1092
edited 21 mins ago
David Z♦
62.3k23135251
edited 21 mins ago
David Z♦
62.3k23135251
edited 21 mins ago
David Z♦
62.3k23135251
62.3k23135251
asked 5 hours ago
SuperMage1
1092
asked 5 hours ago
SuperMage1
1092
asked 5 hours ago
SuperMage1
1092
1092
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
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.
answered 4 hours ago
Pieter
6,09931127
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
@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
4 hours ago
1
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
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
2 hours ago
 |Â
show 4 more comments
up vote
3
down vote
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:
$$L_I=10 Log_10(frac II_0)dB$$
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
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
4 hours ago
1
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
3 hours ago
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
1 hour ago
add a comment |Â
up vote
1
down vote
As already mentioned, decibels are on a logarithmic scale:
$$L=10log_10left(fracII_0right) rmdB$$
If we do what the answer says: increase $I$ by a factor of $10$, what happens?
$$L'=10log_10left(frac10II_0right) rmdB=10log_10left(fracII_0right) rmdB+10 rmdB$$
So we see that we actually add $10 rmdB$ 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
$$2L=10times2log_10left(fracII_0right) rmdB=10log_10left(fracI^2/I_0I_0right) rmdB$$
So you would want your intensity to become
$$I'=fracI^2I_0$$ (i.e., increase by a factor of $fracII_0$)
Which does in fact depend on the initial intensity $I$.
So really either the question could have been written better (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 $10$.
answered 3 hours ago
Aaron Stevens
4,1081624
1
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
3 hours ago
@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
3 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
 |Â
show 2 more comments
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
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.
answered 4 hours ago
Pieter
6,09931127
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
@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
4 hours ago
1
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
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
2 hours ago
 |Â
show 4 more comments
up vote
6
down vote
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.
answered 4 hours ago
Pieter
6,09931127
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
@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
4 hours ago
1
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
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
2 hours ago
 |Â
show 4 more comments
up vote
6
down vote
up vote
6
down vote
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.
answered 4 hours ago
Pieter
6,09931127
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.
answered 4 hours ago
Pieter
6,09931127
edited 4 hours ago
answered 4 hours ago
Pieter
6,09931127
answered 4 hours ago
Pieter
6,09931127
answered 4 hours ago
Pieter
6,09931127
6,09931127
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
@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
4 hours ago
1
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
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
2 hours ago
 |Â
show 4 more comments
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
@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
4 hours ago
1
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
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
2 hours ago
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
@AaronStevens So how would you answer that question in Giancoli?
– Pieter
4 hours ago
2
2
@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
4 hours ago
@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
4 hours ago
1
1
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
4 hours ago
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
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
This doesn't help the OP. If you were to give him another question like "what do we do to make the "loudness" (dB) 5 times as loud?" the OP would not understand, You have essentially just made an answer of "because that is how it is". Hence the downvote. I did not down vote you because nothing is correct here.
– Aaron Stevens
4 hours ago
1
1
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
2 hours ago
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
2 hours ago
 |Â
show 4 more comments
up vote
3
down vote
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:
$$L_I=10 Log_10(frac II_0)dB$$
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
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
4 hours ago
1
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
3 hours ago
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
1 hour ago
add a comment |Â
up vote
3
down vote
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:
$$L_I=10 Log_10(frac II_0)dB$$
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
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
4 hours ago
1
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
3 hours ago
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
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
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:
$$L_I=10 Log_10(frac II_0)dB$$
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
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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:
$$L_I=10 Log_10(frac II_0)dB$$
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
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
answered 5 hours ago
Hugo V
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
Hugo V
715
answered 5 hours ago
Hugo V
715
715
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hugo V is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
4 hours ago
1
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
3 hours ago
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
1 hour ago
add a comment |Â
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
4 hours ago
1
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
3 hours ago
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
1 hour ago
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
4 hours ago
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
4 hours ago
1
1
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
3 hours ago
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
3 hours ago
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
1 hour ago
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
1 hour ago
add a comment |Â
up vote
1
down vote
As already mentioned, decibels are on a logarithmic scale:
$$L=10log_10left(fracII_0right) rmdB$$
If we do what the answer says: increase $I$ by a factor of $10$, what happens?
$$L'=10log_10left(frac10II_0right) rmdB=10log_10left(fracII_0right) rmdB+10 rmdB$$
So we see that we actually add $10 rmdB$ 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
$$2L=10times2log_10left(fracII_0right) rmdB=10log_10left(fracI^2/I_0I_0right) rmdB$$
So you would want your intensity to become
$$I'=fracI^2I_0$$ (i.e., increase by a factor of $fracII_0$)
Which does in fact depend on the initial intensity $I$.
So really either the question could have been written better (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 $10$.
answered 3 hours ago
Aaron Stevens
4,1081624
1
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
3 hours ago
@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
3 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
 |Â
show 2 more comments
up vote
1
down vote
As already mentioned, decibels are on a logarithmic scale:
$$L=10log_10left(fracII_0right) rmdB$$
If we do what the answer says: increase $I$ by a factor of $10$, what happens?
$$L'=10log_10left(frac10II_0right) rmdB=10log_10left(fracII_0right) rmdB+10 rmdB$$
So we see that we actually add $10 rmdB$ 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
$$2L=10times2log_10left(fracII_0right) rmdB=10log_10left(fracI^2/I_0I_0right) rmdB$$
So you would want your intensity to become
$$I'=fracI^2I_0$$ (i.e., increase by a factor of $fracII_0$)
Which does in fact depend on the initial intensity $I$.
So really either the question could have been written better (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 $10$.
answered 3 hours ago
Aaron Stevens
4,1081624
1
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
3 hours ago
@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
3 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
 |Â
show 2 more comments
up vote
1
down vote
up vote
1
down vote
As already mentioned, decibels are on a logarithmic scale:
$$L=10log_10left(fracII_0right) rmdB$$
If we do what the answer says: increase $I$ by a factor of $10$, what happens?
$$L'=10log_10left(frac10II_0right) rmdB=10log_10left(fracII_0right) rmdB+10 rmdB$$
So we see that we actually add $10 rmdB$ 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
$$2L=10times2log_10left(fracII_0right) rmdB=10log_10left(fracI^2/I_0I_0right) rmdB$$
So you would want your intensity to become
$$I'=fracI^2I_0$$ (i.e., increase by a factor of $fracII_0$)
Which does in fact depend on the initial intensity $I$.
So really either the question could have been written better (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 $10$.
answered 3 hours ago
Aaron Stevens
4,1081624
As already mentioned, decibels are on a logarithmic scale:
$$L=10log_10left(fracII_0right) rmdB$$
If we do what the answer says: increase $I$ by a factor of $10$, what happens?
$$L'=10log_10left(frac10II_0right) rmdB=10log_10left(fracII_0right) rmdB+10 rmdB$$
So we see that we actually add $10 rmdB$ 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
$$2L=10times2log_10left(fracII_0right) rmdB=10log_10left(fracI^2/I_0I_0right) rmdB$$
So you would want your intensity to become
$$I'=fracI^2I_0$$ (i.e., increase by a factor of $fracII_0$)
Which does in fact depend on the initial intensity $I$.
So really either the question could have been written better (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 $10$.
answered 3 hours ago
Aaron Stevens
4,1081624
edited 1 hour ago
answered 3 hours ago
Aaron Stevens
4,1081624
answered 3 hours ago
Aaron Stevens
4,1081624
answered 3 hours ago
Aaron Stevens
4,1081624
4,1081624
1
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
3 hours ago
@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
3 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
 |Â
show 2 more comments
1
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
3 hours ago
@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
3 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
1
1
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
3 hours ago
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
3 hours ago
@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
3 hours ago
@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
3 hours ago
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
2 hours ago
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
2 hours ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
@Pieter I know this. I am going off of what the OP was mentioning about trying to double the decibels. I already mentioned in my answer that loudness is subjective. This is why I said double decibels.
– Aaron Stevens
1 hour ago
1
1
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
@Pieter This is true. Thanks
– Aaron Stevens
1 hour ago
 |Â
show 2 more comments
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