Tympani in a concert hall: calculating Sound Pressure Level (SPL) and delay

[Rorshach]

Homework Statement

Hello. I have a problem, which is as follows:
A seat in a concert hall is 84 ft from the tympani. The tympanist strikesa single note. The sound pressure level of the direct sound of the note at the seat is measured to be 55 dB. The first reflection from the nearest sidewall arrives at the seat 105 msec after the arrival of the direct sound.
A) How far does the reflection travel to reach the seat?
B) What is the SPL of the reflection at the seat, assuming perfect reflection at the wall?
C) How long will the reflection be delayed after arrival of the direct sound at the seat?

Homework Equations

sound speed in air:
1130 ft/sec
SPL=20log(p/p_ref)

The Attempt at a Solution

The thing is, they are all solved in the book right under the data, but I don't understand anything beyond the first question and its solution. I would like to ask for some explanation on how why this is solved in this particular way.
First one is easy, I got this:

A)distance=(1130ft/sec)(0.105 sec)=118.7 ft

Please help me understand the other two questions.
Solutions to the are:
B)52 dB
C)30.7 msec

So far point C) is most confusing to me, isn't the answer given in the problem(105 msec)?

 

[haruspex]

B)52 dB

How does sound pressure depend on distance?

A)distance=(1130ft/sec)(0.105 sec)=118.7 ft

That is wrong. If this is given as the correct answer then I conclude that the information given is garbled. My guess is that the given 105msec is not the delay after the direct sound. That would explain why question C makes no sense.

 

[Rorshach]

I wrote everything down word for word from my book. I am barely on page 30 and so far there have been few questionable equations, but those two just boggle me. Under every question there are equations and small adnotations for what is being done to solve the problem, so I will write entire problem as it stands in the book, starting with solution for the first question and ending on the last question:

A) How far does the reflectiontravel to reach the seat?
Distance=(1130 ft/sec)(0.105 sec)=118.7 ft

B)What is the SPL of the reflection at the seat, assuming perfect reflection at the wall?
First, the level L, 1 ft from the tympani, must be estimated:
55=L-20log(84/1)
L=55+38.5=93.5dB

The SPL of the reflection at the seat is:
dB=93.5-20log(118.7/1)=93.5-41.5=52dB

C)How long will the reflection be delayed after arrival of the direct sound at the seat?
The reflection will arrive after the direct sound at the seat after:
delay=(118.7-84)/1130ft/sec=30.7 msec
A free field is also assumed here. The 30.7 msec reflection might be called an incipient echo.
I am lost on how this is done, please people, if you understand how this is solved like this- explain it to me like to a 6 yo.

 

[haruspex]

Suppose it takes time t for the direct sound to arrive.
According to the question, it takes time t+0.105 s for the reflected sound. That makes the total distance traveled by the reflected sound 1130(t+0.105)=84+118.7ft.
Judging from the given answers to a and c, I suggest the question is supposed to read:
"The first reflection from the nearest sidewall arrives at the seat 105 msec after being emitted"

 

[Rorshach]

I am sorry, but I still don't understand it. Maybe one of my brain cells is locked, but the the second, and especially the third question and their solutions are not understandable to me. Why the "delay" is not 105 msec? That question is constructed in such a way that it leaves no option but to think that 105 msec is what they ask for.

 

[haruspex]

Why the "delay" is not 105 msec?

You do not seem to understand my previous post. I am saying that the question has been misstated. It is supposed to say that the 105ms is from when the timpani made the sound to when the echo arrived. That makes the answer 118.7ft correct for a), and the time between the two arrivals is 105ms-(84/1130)s = 31msec.

I will reply separately re b).

 

[haruspex]

For b, how does the sound pressure depend on distance?

 

[Rorshach]

Sorry for not replying for so long- my laptop had a severe malfunction of graphics card after windows 10 update. Sound pressure decreases with 1/r at a distance from sound source.

 

[haruspex]

Sorry for not replying for so long- my laptop had a severe malfunction of graphics card after windows 10 update. Sound pressure decreases with 1/r at a distance from sound source.

Ok.
You are given the SPL of the direct sound. If the sound pressure of the direct sound is p_direct, what is the relationship between that and p_ref?
Using the known distances, what is the relationship between p_direct and p_reflected?

 

[Rorshach]

I understood it some time ago, sorry for not responding for so long- did You mean this approach:
p ∝ 1/r
p₁ ∝ 1/r₁
p₂ ∝ 1/r₂
p₂/p₁ = r₁/r₂
p₂ = p₁ * r₁/r₂

SPL_reflected = 20log₁₀(p_reflected/p_reference)

 

[haruspex]

I understood it some time ago, sorry for not responding for so long- did You mean this approach:
p ∝ 1/r
p₁ ∝ 1/r₁
p₂ ∝ 1/r₂
p₂/p₁ = r₁/r₂
p₂ = p₁ * r₁/r₂

SPL_reflected = 20log₁₀(p_reflected/p_reference)

Yes, except that that sequence is not mathematically correct.
p₁ ∝ 1/r₁ does not mean anything. Proportionality is a property of variables, whereas p₁ and p₂ are constants. Better would be
p ∝ 1/r
p =k/r for some constant k
p₁ =k/r₁
Etc.

 

[Rorshach]

Yes, You are right. There is also one more problem- an easy one, about loudspeaker SPL.In the book they give the solution, but when I checked it on calculator, the results were different. Results they gave were intuitive, but formulas they used give completely different result. Do You think I can post this problem in this thread or should I start a new one?

 

[haruspex]

Yes, You are right. There is also one more problem- an easy one, about loudspeaker SPL.In the book they give the solution, but when I checked it on calculator, the results were different. Results they gave were intuitive, but formulas they used give completely different result. Do You think I can post this problem in this thread or should I start a new one?

Better to post a new one.

 

[Rorshach]

Ok, I will. Thank You for your help- I have one question regarding the approach You suggested. The results from method proposed in the book (52 dB) differ slightly from the one You proposed(~51dB). Is it ok? I mean, the dependence of acoustic pressure on distance is not a strict rule, but more of an approximation, right?

 

[haruspex]

the one You proposed(~51dB).

I get 52dB. Please post your working.

 

[Rorshach]

55 dB = 20log(p₁/p_ref)
55 = 20log(p₁/20*10^-6)
2.75 = log(p₁/20*10^-6)
10^2.75 = p₁/20*10^-6
p₁ = 10^2.75 * 20*10^-6
p₁ = 0.01 Pa

r₁ = 84 ft
r₂ = 118.7 ft

p∝1/r
p₁ = k/r₁
p₂ = k/r₂
p₂ = p₁ * r₁/r₂p₂ = p₁ * r₁/r₂
p₂ = 0.01 Pa * 84/118.7 = 0.007 Pa

SPL₂ = 20log(0.007/(20*10^-6)) = 50.881 ≈ 51 dB

 

[haruspex]

p1 = 0.01 Pa

Here you are down to only 1 sig fig.
By plugging in numbers straight away you run the risk of a lot of rounding error accumulation.
There are many advantages in keeping everything symbolic until the end.

r₁ = 84ft
t₂ = 0.105s
c = 1130f/s
r₂ = ct₂
p₁/p₂=r₂/r₁
SPL₂=SPL₁+20 log(p₂/p₁)
= SPL₁+20 log(r₁/(ct₂))
= 55+20 log(84/(1130x0.105)) = 52