i) a metre long tube open at one end with a movable position at the other and shows resonance with a fixed frequency source ( a tuning fork of frequency 340Hz) when this tube length is 25.5 CM or 70 9.3 CM. estimate the speed of sound in the air at the temperature of the experiment. the adge effects may be neglected.....
Answers
Answer:
Let ..V...be the fundamental frequency of the closed and organ pipe for the length
L* = 25.5 cm.
Let 'u' be the speed of sound,..
Then,
V. = u / 4 L
Now, Frequency of the n'th mode of vibration of close end pipe for length L*
V* = ( 2n - 1 ) v
= ( 2n - 1 ) u / 4 L*.............(1)
Also, The frequency of the (n +1)th mode of vibration of close end pipe for length l° = 79.3 cm is...
V° = [ 2 (n - 1 ) - 1 ] u / 4 L°
= ( 2n + 1 ) u / 4 L°….......…………(2)
Since the n'th mode and (n+1) mode are in resonance with the frequency of 340Hz,
V* = V°
=⟩⟩....
( 2n - 1 ) u / 4 L*. = ( 2n + 1 ) u /4L°
Or...2n - 1 / 2n + 1 = L* / L °
= 25.5 / 79.3 = 0.32
=⟩⟩ ....
2n - 1 = 0.32 (2n + 1) = 0.64n + 0.32
Or. 2n - 0.64n = 1 + 0.32
=⟩⟩
Becoz n is a whole number...so....
1.36n = 1.32 ......( n ~ 1 )
Using this value in (1)...
V* = ( 2n - 1 ) u / 4 L°
=⟩⟩. 340 = (2 x 1 - 1) x u / 4 x 25.5
=⟩⟩. u = 340 x 4 x 25.5 cm/s
= 34680cm/s
= 346.8 m/s....