Max velocity of a vibrating loud speaker membrane given sound intensity

AI Thread Summary
The discussion centers on calculating the maximum velocity of a vibrating loudspeaker membrane using sound intensity and related formulas. The speaker's diaphragm is noted to be 1mm in diameter, and the sound intensity at 10 meters is calculated to be 10^-7 W/m^2. There is confusion regarding the applicability of the formula I=0.5*Z*v_max^2 for spherical versus planar waves, with a suggestion that the speaker's output should be analyzed using the inverse square law for intensity. The conversation also touches on the efficiency of the speaker based on its mounting and the implications for sound wave propagation. Overall, the calculations and assumptions about wave behavior and speaker design are critical to understanding the results.
custner
Messages
3
Reaction score
0
Homework Statement
This for my first university course in waves.

p=105 kPa and T= 300 K. We are in air. A loudspeaker with a circular membrane of diameter 1 mm sends out sound. At a distance of 10 meters, the sound intensity level is 50 dB. What is the maximum velocity that the loudspeakers membrane vibrates with?
Relevant Equations
Z=p*sqrt(gamma*M/(R*T))
dB = 10*log(I/I_0)
I=0.5∗Z∗(v_(max))^2
My attempt:
p and T allows us to calculate ##Z=402 \frac{kg}{sm^2}## using ## Z=p*\sqrt(\frac{\gamma*M}{R*T})## . The sound intensity level at 10 meters allows us to calculate the intensity at 10 meters to be I=10``````^{-7} W/m^2 using ##50 = 10*log(I/I_0)##. Then, using the formula ##I=0.5∗Z∗v_{max}^2##, which gives ##v_{max}=2.23∗10^{−5} m/s##My question:
Sound waves are spherical waves, but the expression ##I=0.5∗Z∗v_{max}^2## is (from what I understand) for planar waves. This makes me think that it is incorrect to use it since sound waves (I think?) are spherical waves. Is it still correct to use it for sound waves? Because the answer to me feels very small
 
Last edited:
Physics news on Phys.org
I can't seem to edit the question again. But I think that the v_max I get is for the wave at 10 meters. So somehow I need to find the intensity at the membrane right? But using the formula I_1*r_1^2=I_2*r_2^2 with r_1=10 meters would give be I_2=infinity sine r_2=0, or am I missing something?
 
Last edited:
I got a hint today that that solution should use conservation of energy, but I don't know how that helps
 
Well, you haven’t got any replies and I can’t follow some of what you have written. Also a 1mm diameter speaker sounds unrealistic - are you sure this is correct?

But I feel some sympathy (!) so here’s a possible approach...

1. Work out the intensity (I) at 10m in ##W/m^2##.

2. If there are no energy losses, and the sound distribution is spherically symmetric, then the speaker’s power output (P, in watts) is related to the intensity$$I = \frac{P}{4πr^2}$$ where r=10m. This is simply using the inverse square law. Find P.

3. Now use formula 4.70 here: https://www.sciencedirect.com/topics/engineering/radiated-sound-power Note the link uses ‘W’ rather than ‘P’ for power and you can assume efficiency (σ) is 1 or estimate its value somehow.

You may want to find the speed of sound and density for your given temperature and pressure, though I would have thought ball-park figures are OK for this sort of calculation.

You simply work out the rms speed, ##√ <v^2>##

4. Multiply by ##√2## (assuming speaker is performing SHM) to find max. (peak) speed.

I've given you much more guidance than I should, so I hope the rest of the community will forgive me!

You will find using Latex for formulae makes your posting clearer and makes readers more likely to reply.
 
custner said:
My question:
Sound waves are spherical waves, but the expression ##I=0.5∗Z∗v_{max}^2## is (from what I understand) for planar waves. This makes me think that it is incorrect to use it since sound waves (I think?) are spherical waves. Is it still correct to use it for sound waves? Because the answer to me feels very small
Your speaker diaphragm is 1mm in diameter and you are measuring 10 metres away.

The waves will be launched as planar waves, and be planar very, very close to the diaphragm (a millimetre?) but will become approximately spherical very, very quickly.

Don't expect any base frequencies :smile:

EDIT: A thought. If the diaphragm is free standing then the high pressure at the front when it moves forwards will tend to be canceled by the low pressure at the back as air can flow round the edge of the diaphragm. It is why speakers are mounted on baffles and the effect is more pronounced the lower the frequency.

So, either it is not mounted on a baffle, in which case it will be very inefficient (especially at low frequencies); or it is mounted on a baffle, in which case it will be directional.

Ho-hum!
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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