One end of A 20 cm long tube is closed. Find the harmonic mode of the tube that will be resonantly excited by a source of frequency 430 Hz. lf both the ends are open, can the same source still produce resonance in the tube?(Sound travels in air at 340m/s).

Asked by Pragya Singh | 1 year ago |  108

##### Solution :-

Given,
Length of the pipe, l = 20 cm = 0.2 m
Frequency of the source = nth
the normal mode of frequency, νN = 430 Hz
Speed of sound, v = 340 ms -1

We know, that in a closed pipe the nth normal mode of frequency

v $$\dfrac{ ( 2n -1 )v}{4I}$$
where n is an integer = 0, 1, 2, 3, 4, . . . . .

430 = ( 2n – 1 )($$\dfrac{340}{4}\times 0.2$$ )
2n = 2.01
n ≈ 1
Thus, the given source resonantly excites the first mode of vibration frequency

Now, for a pipe open at both the ends, the nth  mode of vibration frequency:

VR = $$\dfrac{ nv}{2l}$$

n = $$\dfrac{ V_R 2I}{v}$$

n = $$\dfrac{ (2 \times 0.2 \times 430)}{340}$$ = 0.5

As the mode of vibration ( n ) has to be an integer, this source is not in resonance with the tube.

Answered by Pragya Singh | 1 year ago

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