Answer :
Answer:
(a) n = 6
(b) N = 7
Solution:
As per the question:
Length of the tube, [tex]L_{m} = 4\ m[/tex] = 4 m
Frequency, [tex]\nu = 398\ Hz[/tex]
Speed of sound, v = 334. m/s
Now,
(a) To calculate the no. 'n':
We know that:
[tex]\nu = v\lambda[/tex]
Thus
[tex]\lambda = \frac{\nu}{v} = \frac{398}{334.4} = 1.1902\ m[/tex]
Now,
[tex]L_{m} = \frac{2n_{m} + 1}{4}\lambda [/tex]
[tex]4 = \frac{2n_{m} + 1}{4}\times 1.1902[/tex]
[tex]n_{m} = 6[/tex]
(b) The no. of resonance observed is given by:
[tex]N = n_{m} + 1 = 6 + 1 = 7[/tex]