Pressure Amplitude Calculation for Sound Waves

AI Thread Summary
To calculate the pressure amplitude of a sound wave with an intensity level of 65 dB, the relationship between intensity and pressure amplitude needs to be understood. The user attempts to derive pressure from intensity using the formula PRESSURE = I / V_SOUND, where V_SOUND is approximately 340 m/s. However, the calculations do not yield the expected result according to the answer key. The discussion highlights a lack of clarity in the textbook regarding the connection between intensity levels in decibels and pressure amplitude. Further guidance or clarification on the correct formula for pressure amplitude in relation to intensity levels is sought.
akan
Messages
58
Reaction score
0

Homework Statement


What is the pressure amplitude in sound wave with intensity level of 65 dB?


The Attempt at a Solution


My textbook does not provide any instruction on how pressure amplitude is related to the intensity level. It does not even define what pressure amplitude is. While I can calculate the intensity (P/A) from decibels, I do not know how to calculate pressure amplitude. Please help.
 
Physics news on Phys.org
Ok, so here's my attempt at solution..
I = POWER/AREA = FORCE * VELOCITY / AREA
PRESSURE = FORCE/AREA
PRESSURE = I / VELOCITY.
Since we are talking about sound, I suppose it is PRESSURE = I / V_SOUND (340), where I calculated I to be (I_0) (10^6.5).
But this doesn't seem to give the right answer (according to the answer key). What's the issue here?
 
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}...

Similar threads

Back
Top