This is a problem that we can solve scientifically. And it will require collaboration.
And it's kind of fun o.
Here it is.

how many people are required to form a choir that that will deafen everyone singing in it? Assuming they are put in a relatively closed environment? Such as a church?

Biologists, physicists, and mathematicians are needed. Let's see if we can solve this.

Cc robinhez johnydon22 richiez faba umartins1 venus007 dollydak feraz electgineer dekatron saintneo pukka36 and so so on.

Okay . . . let me kick this off. . .

With some research, I found that an average sound of 145 decibels is required to deafen a human. However, the voice of humans shouting is just around 88 decibels.

Let us assume that each person is singing at a sound intensity of 85 decibels.

In that case, we can interpret the question as, how many sound sources vibrating at 85 decibels is enough to produce a sound of 145 decibels ?

IMHO, it will take around 5 people to do so, shouting at the top of their voices. But it all depends on the location. Because of the nature of sound waves, dome of the waves are bound to move out to space and get lost while some are bound to be reflected by an object. It all depends on the area or location of the scene; Its different if it were fine in a closed space or inside a room. Its also different if it were fine in an open space.

Pakingzzz:
IMHO, it will take around 5 people to do so, shouting at the top of their voices. But it all depends on the location. Because of the nature of sound waves, dome of the waves are bound to move out to space and get lost while some are bound to be reflected by an object. It all depends on the area or location of the scene; Its different if it were fine in a closed space or inside a room. Its also different if it were fine in an open space.

what if we assume that they are in a closed studio that doesn't let sound escape? Would the number increase or decrease?

Teempakguy:
what if we assume that they are in a closed studio that doesn't let sound escape? Would the number increase or decrease?

The number of people needed would decrease.

U already gave the solution!

Maybe I should read sound waves, I only love electricity.

Teempakguy:
Okay . . . let me kick this off. . .

With some research, I found that an average sound of 145 decibels is required to deafen a human. However, the voice of humans shouting is just around 88 decibels.

Let us assume that each person is singing at a sound intensity of 85 decibels.

In that case, we can interpret the question as, how many sound sources vibrating at 85 decibels is enough to produce a sound of 145 decibels ?

TWO people producing an average of 85 decibels each is enough... That's if they are a contained environment i.e an anechoic chamber

I think factors that affect sound waves should be considered. nature of surrounding objects & stuffs


Anyway, you have it. 145/85 . Thats all. I think

When all of una don deceive una sef finish make una come tell me...


Have u all not heard dat music is different from noise E dey Essential Physics.

And music simply put is wen sounds agree and are produced smootly as one...if una still no get me...think of heaven countly angels singing harmoneously to the Father of Light and no one goes deaf...

So you guys mean we should just divide?

I don't support that logic at all oh.

Consider a speaker put in front of a person. It's a meter tall, and it's spitting badass music, which by the way, is usually around 100decibels.
Are you saying that if we bring a similar speaker and put it in front of the person, we will get a sound of 200 decibels? Which, by the way, is the sound of a VERY VERY BIG bomb explosion.

anyway, I found a formula for adding decibels.
If we have a decibel value x, then,

2x = x + 3

This implies something like 88 decibels, is actually double of 85 decibels.

By adding 3 decibels, you increase the number of people singing. By doubling them. So, now, if you keep doing that, when will you get to 145?

Okay guys, na wa o, but you guys have tricked me into solving both the biological and physics aspect! Smh . . . Nigerians are smart sha . . . anyway, oyah, it remains the mathematical part, who can solve it?

Dekatron robinhez pakingzzz richiez and er . . . what other maths guru is there oh . . . umartins1 . . .

Teempakguy:
So you guys mean we should just divide?

I don't support that logic at all oh.

Consider a speaker put in front of a person. It's a meter tall, and it's spitting badass music, which by the way, is usually around 100decibels.
Are you saying that if we bring a similar speaker and put it in front of the person, we will get a sound of 200 decibels? Which, by the way, is the sound of a VERY VERY BIG bomb explosion.

anyway, I found a formula for adding decibels.
If we have a decibel value x, then,

2x = x + 3

This implies something like 88 decibels, is actually double of 85 decibels.

By adding 3 decibels, you increase the number of people singing. By doubling them. So, now, if you keep doing that, when will you get to 145?

Okay guys, na wa o, but you guys have tricked me into solving both the biological and physics aspect! Smh . . . Nigerians are smart sha . . . anyway, oyah, it remains the mathematical part, who can solve it?

Dekatron robinhez pakingzzz richiez and er . . . what other maths guru is there oh . . . umartins1 . . .

there are a lot more factors to be put into consideration...you don't expect a soprano singer, an alto singer and a bass singer to all produce the same decibel of sound...consider that too

All the answers given here are all too simple and easy. to actually solve this, you need a solid understanding of acoustic engineering.

besides, I'd like to chip in the fact that sound wave deteriorates over distance. so basically, the sound generated 10 meters away and of one generated 50m away irrespective of the acoustic insulation of the environment would not be the same in intensity.

cupid1:
there are a lot more factors to be put into consideration...you don't expect a soprano singer, an alto singer and a bass singer to all produce the same decibel of sound...consider that too

that would be a highly complex model. So, we are going to have to assume that they are all singing the same tone.

It's possible nah.

destino24:
All the answers given here are all too simple and easy. to actually solve this, you need a solid understanding of acoustic engineering.

besides, I'd like to chip in the fact that sound wave deteriorates over distance. so basically, the sound generated 10 meters away and of one generated 50m away irrespective of the acoustic insulation of the environment would not be the same in intensity.

a choir huddled around themselves would act as a single sound source. Since we are asking about what would happen within that sound source, we don't need to calculate for sound deterioration over distance.

Acoustic engineering, yes. But this is called a "back of the envelope question" where you source for needed values and solve for rough values. It's usually done for fun among scientists and is never intended to be highly correct.

This is interesting.. lemme just learn

Teempakguy:
a choir huddled around themselves would act as a single sound source. Since we are asking about what would happen within that sound source, we don't need to calculate for sound deterioration over distance.

Acoustic engineering, yes. But this is called a "back of the envelope question" where you source for needed values and solve for rough values. It's usually done for fun among scientists and is never intended to be highly correct.

back of the envelop, huh?
OK.

after a little research I realized that an increase in decibel units by 10, in reality is an increase in sound x10.
Humans produce roughly 90 db when shouting, to get to 100db, the human has to shout 10 times as much.
So, assuming 150db would deafen a human,

90db = I human sound
100db = 1 human sound x 10= 10 humans
110db = 10 * 10 = 100
120db = 100*10=1000
130db = 1000*10=10000
140db = 10000*10=100000
150db = 100000*10=1,000,000 humans

The funny analogy up there assumes that sonic deterioration did not take place, sonic insulation of the environment is working at 100 percent efficiency and that all the humans singing are all "singing" at the same intensity, tone and at 90db among other numerous factors

destino24:


back of the envelop, huh?
OK.

after a little research I realized that an increase in decibel units by 10, in reality is an increase in sound x10.
Humans produce roughly 90 db when shouting, to get to 100db, the human has to shout 10 times as much.
So, assuming 150db would deafen a human,

90db = I human sound
100db = 1 human sound x 10= 10 humans
110db = 10 * 10 = 100
120db = 100*10=1000
130db = 1000*10=10000
140db = 10000*10=100000
150db = 100000*10=1,000,000 humans

The funny analogy up there assumes that sonic deterioration did not take place, sonic insulation of the environment is working at 100 percent efficiency and that all the humans singing are all "singing" at the same intensity, tone and at 90db among other numerous factors

see? Now we know that anything more than at least 200 000 humans singing in a cathedral at a high voice would be very dangerous to each person. So, when you read something like.
"Biggest church in the world houses over 200 000 people, you would be able to ask an informed question that, how the hell do they manage the sound?" Which would lead you to study the architecture of the place, and learn something else in return!

And that's really the beauty of science. It doesn't have to be correct, it just has to work for what we need it too.

Anyway, thanks for solving this.

Teempakguy:
that would be a highly complex model. So, we are going to have to assume that they are all singing the same tone.

It's possible nah.

then you'll have to change your case study...use something else instead of a choir 'cos no choir sings d same note

cupid1:
then you'll have to change your case study...use something else instead of a choir 'cos no choir sings d same note

we have an average of 1 000 000 people that will deafen the individuals. Do you still think that slight variations in frequency can seriously change that?

Teempakguy:
we have an average of 1 000 000 people that will deafen the individuals. Do you still think that slight variations in frequency can seriously change that?

i do....don't you? It might end up being 9,999,997 or 1,000,003

cupid1:
i do....don't you? It might end up being 9,999,997 or 1,000,003

yeah, it could also be 9 999 999.7689
Besides, how do you know which percentage is singing a particular tone?

You're building up a very complicated model there man, can you solve it?

Teempakguy:
yeah, it could also be 9 999 999.7689
Besides, how do you know which percentage is singing a particular tone?

You're building up a very complicated model there man, can you solve it?

why that 1,000,000 is quite erronous is the fact that even in real life, no two people sing the same note, on the same pitch. And if you start calculating for different notes, you'd have to factor in things pitch and the other little knik-knacks that make up voices. And that's a whole lot of odds
.
.
.
.
.
So in real life that calculation is impossible...only possible in physics where generalizations are allowed

Teempakguy:
So you guys mean we should just divide?

I don't support that logic at all oh.

Consider a speaker put in front of a person. It's a meter tall, and it's spitting badass music, which by the way, is usually around 100decibels.
Are you saying that if we bring a similar speaker and put it in front of the person, we will get a sound of 200 decibels? Which, by the way, is the sound of a VERY VERY BIG bomb explosion.

anyway, I found a formula for adding decibels.
If we have a decibel value x, then,

2x = x + 3

This implies something like 88 decibels, is actually double of 85 decibels.

By adding 3 decibels, you increase the number of people singing. By doubling them. So, now, if you keep doing that, when will you get to 145?

Okay guys, na wa o, but you guys have tricked me into solving both the biological and physics aspect! Smh . . . Nigerians are smart sha . . . anyway, oyah, it remains the mathematical part, who can solve it?

Dekatron robinhez pakingzzz richiez and er . . . what other maths guru is there oh . . . umartins1 . . .

i think this would require careful experimentation and ADVANCED MATH -- CALCULUS? THEORY OF WAVES? COMBINATION OF BOTH IN THEIR ADVANCED STATE? DUNNO.

But this, if we can't divide would need more time and math to battle.


Study the relationship between the factors. Thats the math, i GUESS. And that can be achieved by : EXPERIMENTING