Determining the sound level of a leaf blower

  • Thread starter Thread starter barryj
  • Start date Start date
  • Tags Tags
    distance Sound
Click For Summary
The sound level of a leaf blower rated at 75dB is typically measured at a distance of 50 feet, as indicated by a manufacturer's lobbyist. Sound intensity increases as one gets closer to the source, with a doubling of distance resulting in a decrease of approximately 6 dB. This means that at closer ranges, the perceived noise level can be significantly higher. The discussion highlights the importance of understanding measurement distances when evaluating noise levels. Accurate sound level assessments are crucial for compliance with noise regulations and for user comfort.
barryj
Messages
856
Reaction score
51
Homework Statement
How does one determine the dB of a leaf blower.
Relevant Equations
db = 10log(p1/p2)
I have a leaf blower that states the sound level is 75dB. My question is, at what distance is this measured. It seems that as you get closer to the source, the dB number will increase drastically. If I double the distance from the source, then the power should go down by a factor of four or 6 db. Is this correct?
 
Physics news on Phys.org
You are correct. Can you post a link for the product?
If not stated, I would take it as the expected distance from the device to the operator's ears.

Update: found this
https://www.nonoise.org/quietnet/cqs/leafblow.htm
"The average blower measures 70-75 dB at 50 feet according to a manufacturer's lobbyist"
So maybe 50 feet is standard.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
View full post »

Similar threads

Measuring the Velocity of Sound in this Lab Exercise
Replies
19
Views
2K
Sound Power Level (SPL) of a loudspeaker
Replies
11
Views
2K
I Sound power, amplitude, frequency, and decibels
Replies
5
Views
2K
Attenuation Coefficient in an optical fibre
Replies
21
Views
2K
Calculating Sound Intensity and Pressure Variation at a Distance
Replies
4
Views
2K
Centrifugal blower ringing?
Replies
22
Views
3K
Calculating Power at source from Intensity at distance R
Replies
1
Views
5K
Sound Wave Interference and Finding the Path Differences with Diagrams
Replies
20
Views
5K
Change in sound intensity over distance
Replies
1
Views
4K
Problem on Doppler effect and sound level
Replies
1
Views
2K