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# A tuning fork while vibrating with an air column at 51 degree celcius produces 4 beats and the same tuning fork produces 1 beats at 16 degree celcius. Find Frequency Of Tuning Fork. Ans 50Hz

Ramesh V
70 Points
11 years ago

when temp. decreases, speed of sound in air decreases, since no. of beats per second is less at lower tempertaure we coclude that frequency of aircolumn is higher than freq. of tuning fork.

let v be freq. of tuning fork, then at 51 degrees frq. of air column will be (n+4) where n is freq. of tuning fork

at 16 degrees , frewq. of air column will be (n+1)

v51/v16 = (n+4)*k / (n+1)*k
(n+4)/(n+1) = (273+51)1/2 / (273+16)1/2

on solving ofr n , we'll get n = 50 Hz

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regards

Ramesh

Nehal Wani
21 Points
11 years ago

Thank You very Much Sir.

Rishi Sharma
one year ago
Dear Student,

when temp. decreases, speed of sound in air decreases, since no. of beats per second is less at lower tempertaure we coclude that frequency of aircolumn is higher than freq. of tuning fork.

let v be freq. of tuning fork, then at 51 degrees frq. of air column will be (n+4) where n is freq. of tuning fork

at 16 degrees , frewq. of air column will be (n+1)v51/v16 = (n+4)*k / (n+1)*k
(n+4)/(n+1) = (273+51)1/2 / (273+16)1/2
on solving ofr n , we'll get n = 50 Hz

Thanks and Regards