Max velocity of a vibrating loud speaker membrane given sound intensity

  • #1
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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
 
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Answers and Replies

  • #2
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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?
 
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  • #3
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I got a hint today that that solution should use conservation of energy, but I don't know how that helps
 
  • #4
Steve4Physics
Homework Helper
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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.
 
  • #5
Frodo
Gold Member
201
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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 cancelled 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!
 
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