Calculating τ by knowing I is proportional to A^2

[Cc518]

Homework Statement

A vibrating standing wave on a string radiates a sound wave with intensity proportional to the square of the standing-wave amplitude. When a piano key is struck and held down, so that the string continues to vibrate, the sound level decreases by 8.0 dB in 1.0 s.

What is the string's damping time constant τ ?

Homework Equations

I∝( 2asin(kx))^2
B=10log(I/1*10^-12)

The Attempt at a Solution

From 8dB, I got change in sound intensity is 6.31*10^-12 w/m2
Since the intensity is proportional to the the square of amplitude, the amplitude will decrease by (6.31*10^-12)^1/2
So I got

(6.31*10^-12)^1/2 = 2a - 2a•e^-t/τ =
2a (1- e^-t/τ)

I assumed sin(kx)=1 because we are looking at the greatest amplitude.
t=1s
I don’t know where to go from here as I don’t know what a is.

Since I don’t know what the original sound intensity is, I won’t be able to know the percentage the sound intensity has decreased in order to calculate the percentage the amplitude has decreased.

Any help is appreciated:)

 

[mjc123]

"Decibels" in this context is not an absolute measure of sound intensity (as in " a noise of 100 dB"), but a relative measure. 1 bel (10dB) is an intensity difference of a factor of 10. A decrease of 8 dB means a decrease by a factor of 10^0.8.

 

[Cc518]

Thank you for reply:)
So does that mean amplitude will decrease by a factor of √10^0.8 ?
If so, then e^-t/τ =√10^0.8,
t=1, I got τ=1.08s which is not the right answer:(
Can anyone tell me where I went wrong?
Thank you!

 

[Merlin3189]

... 1 bel (10dB) is an intensity difference of a factor of 10. A decrease of 8 dB means a decrease by a factor of 10^0.8.

What is not quite so clear from this, is that the 8dB is the intensity rather than the amplitude.
As you noted yourself, intensity ∝ square of amplitude. So the amplitude² decreases by 8 dB, which means the amplitude decreases by ... ?

Thank you for reply:)
So does that mean amplitude will decrease by a factor of √10^0.8 ?

So it means A² decreases by a factor of √10^0.8 (Though it may be easier to calculate it slightly differently, depending on your math preferences.)

 

[Cc518]

So the amplitude2 decreases by 8 dB, which means the amplitude decreases by

Amplitude will decrease by √8dB, right?
Then why did you say

it means A2 decreases by a factor of √10^0.8

?

 

[Merlin3189]

No. Not √8 dB.
If A is the ratio of amplitudes and the ratio of intensities is 8 dB,
$$ 8 = 10 log( A^2 ) \ \ ⇒ \ \ 8 = 20 log( A ) $$
$$so A^2 = 10^{0.8} \ \ and \ \ A = 10^{0.4} $$
$$so A^2 = 6.3 \ \ and \ \ A= 2.5 = \sqrt {6.3} $$

 

[haruspex]

So does that mean amplitude will decrease by a factor of √10^0.8 ?

Yes. You can easily simplify that.

If so, then e^-t/τ =√10^0.8,

It decreases by that factor. Is that factor greater or less than 1?

 

[Cc518]

Yes. You can easily simplify that.

It decreases by that factor. Is that factor greater or less than 1?

The factor is less than 1?
A0/At=√10^0.8 and At=e-t/τ A0
Then e-t/τ = 1/√10^0.8
t=1,
I got τ=1.08s which is twice the answer, 0.54s, though, but I don’t see why I have to divide 1.08 by 2.

 

[haruspex]

Then e-t/τ = 1/√10^0.8

yes, that's better

I got τ=1.08s which is twice the answer

Hmmm.. so do I. Looks like a confusion between amplitude and intensity.