Amplitude of sound wave in by 50%

[hemetite]

Any kind soul to help me with this questions...i am stuck

qn. The amplitude of vibration of a certain sound wave is increase by 50%. What is the corresponding increase in the decibel level of the sound?

My attempt..

beta= 10 log ( I/Io)

lets just take I = 2.0 x 10^-7
let just take Io= 1.00 x 10^-12

then 10 log( 2.0 x 10^-7/1.00 x 10^-12) = 10 log (2.0 x 10^5) = 53db

if we increse I by 50 % = (2.0 x 10^-7) x 1.5
= 3.0 x 10^-7

then
10 log( 3.0 x 10^-7/1.00 x 10^-12) = 10 log (3.0 x 10^5) = 54.8 db

so the change of decibel = 54.8db - 53db = 1.8db

correct attemp this question this way??

 

[LowlyPion]

Any kind soul to help me with this questions...i am stuck

qn. The amplitude of vibration of a certain sound wave is increase by 50%. What is the corresponding increase in the decibel level of the sound?

My attempt..

beta= 10 log ( I/Io)

lets just take I = 2.0 x 10^-7
let just take Io= 1.00 x 10^-12

then 10 log( 2.0 x 10^-7/1.00 x 10^-12) = 10 log (2.0 x 10^5) = 53db

if we increse I by 50 % = (2.0 x 10^-7) x 1.5
= 3.0 x 10^-7

then
10 log( 3.0 x 10^-7/1.00 x 10^-12) = 10 log (3.0 x 10^5) = 54.8 db

so the change of decibel = 54.8db - 53db = 1.8db

correct attemp this question this way??

Since the increase is by a factor of 50% you can also express the increase
as simply 10log10(1.5) = 10*(0.176) = 1.76 db

 

[baba89]

I would like to clarify that the decibel level of a sound wave is determined by its intensity, not its amplitude. The intensity of a sound wave is directly proportional to the square of its amplitude. Therefore, increasing the amplitude of a sound wave by 50% would result in a 125% increase in its intensity, not just 50%.

To calculate the corresponding increase in decibel level, we can use the following formula:

β2 = β1 + 10 log (I2/I1)

where β2 is the new decibel level, β1 is the initial decibel level, I2 is the new intensity, and I1 is the initial intensity.

Using your example, if the initial intensity is 2.0 x 10^-7 and we increase it by 50%, the new intensity would be 3.0 x 10^-7. Plugging these values into the formula, we get:

β2 = 53 + 10 log (3.0 x 10^-7/2.0 x 10^-7) = 53 + 10 log (1.5) = 53 + 1.8 = 54.8 db

Therefore, the corresponding increase in decibel level would be 54.8 dB - 53 dB = 1.8 dB, which is the same as your calculation. However, it is important to note that the increase in decibel level is not directly proportional to the increase in amplitude, but rather to the increase in intensity.

I hope this explanation helps to clarify any confusion and aids in your understanding of sound waves and decibel levels. If you have any further questions, please feel free to ask.