A church organ consists of pipes, each open at one end, of different lengths. The minimum length is 30 mm and the longest is 4 m. Calculate the frequency range of fundamental notes. (speed of sound = 340 m s⁻¹). (Ans: 21 Hz to 2833 Hz)



Data Given:


Speed of sound = v = 340 m s⁻¹

Minimum length = l₁ = 30 mm = 0.03 m

Longest length = l₂ = 4 m




To Find:


Frequency range (f₁  = ? and f₂ = ?)


Solution:


The formula for fundamental frequency for closed pipe at minimum length (l = 0.03 m)

f₁ = v4l

putting values

f₁ = 340ms¹40.03m

f₁ = 2833 Hz -----------Ans. (1)



For maximum length (l₂ = 4 m)

f = v4l 

putting values

f = 340ms¹44m

f₁ = 21.25 Hz 

f₁ = 21 Hz -----------Ans. (2)

Hence the frequency range is from 21 Hz to 2833 Hz.



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK    

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.