Question

In resonance tube experiment if V=300 m/s, n=500 Hz, L=125 cm find 1) Maximum order of resonance 2) Maximum no. of resonance 3) Max. and Min. water level kept in resonance condition

Answer 1

Answer:

1) Maximum order of resonance is 7th harmonic

2) maximum number of resonance is N = 4

3) Maximum water level is 110 cm

minimum water level is 20 cm

Explanation:

As we know that the frequency of the sound is

$n = 500 Hz$

speed of the sound is

v = 300 m/s

now we have

$\lambda = \frac{v}{n}$

$\lambda = \frac{300}{500}$

$\lambda = 0.6 m$

now we have

$\frac{2N + 1}{4}\lambda = L$

$\frac{2N + 1}{4}0.6 = 1.25$

$N = 3.66$

so for integral value of N = 3

1) So maximum order resonance is

n = (2N + 1)

now N = 3

so order is n = 7th

so maximum order of the frequency is 7th harmonic

2) As we know that only integral values of N will exist

so we have

N =0, N = 1, N = 2 & N = 3

so total number of resonance is 4

3) for fundamental frequency the water level is maximum

so length of air column is

$L = \frac{\lambda}{4}$

$L = \frac{0.6}{4} = 0.15 m$

So maximum water in the tube is

$h_{max} = 125 - 15 = 110 cm$

For 7th harmonic length of air column is

$L = \frac{7}{4}\lambda$

$L = 1.05 m$

so minimum water in tube is

$h_{min} = 125 - 105 = 20 cm$

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Topic : Resonance tube

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