Amplitude of sound wave in by 50%

AI Thread Summary
The discussion centers on calculating the increase in decibel level resulting from a 50% increase in the amplitude of a sound wave. The initial calculation shows that increasing the intensity from 2.0 x 10^-7 to 3.0 x 10^-7 leads to an increase of approximately 1.8 dB. Another participant suggests a more straightforward method using the formula 10 log10(1.5), which yields an increase of about 1.76 dB. Both methods effectively demonstrate that a 50% increase in amplitude corresponds to a slight increase in decibel level. The calculations confirm the relationship between amplitude and sound intensity in decibels.
hemetite
Messages
50
Reaction score
0
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??
 
Physics news on Phys.org
hemetite said:
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
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top