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Standing Waves in Open Organ Pipe

The velocity ...

Question

The velocity of sound in air is $333 m/s$ . In order to produce the second overtone of frequency $999 Hz$ in it, the length of the open organ pipe will be :

0.5 m

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1 m

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1.5 m

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0.25 m

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Solution

The correct option is A $0.5 m$
In an open organ pipe all the harmonics are present.
Therefore the frequency of second overtone or third harmonic is,
$f_{3} = \frac{3 v}{2 L}$
$\Rightarrow 999 = \frac{3 \times 333}{2 \times L} \Rightarrow L = \frac{1}{2} m = 0.5 m$

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Standing Waves in Open Organ Pipe

Standard XII Physics

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The velocity of sound in air is 333 m/s. In order to produce the second overtone of frequency 999 Hz in it, the length of the open organ pipe will be :

The correct option is A 0.5 mIn an open organ pipe all the harmonics are present. Therefore the frequency of second overtone or third harmonic is, f3=3v2L ⇒999=3×3332×L⇒L=12 m=0.5 m

The velocity of sound in air is $333 m/s$ . In order to produce the second overtone of frequency $999 Hz$ in it, the length of the open organ pipe will be :

The correct option is A $0.5 m$
In an open organ pipe all the harmonics are present.
Therefore the frequency of second overtone or third harmonic is,
$f_{3} = \frac{3 v}{2 L}$
$\Rightarrow 999 = \frac{3 \times 333}{2 \times L} \Rightarrow L = \frac{1}{2} m = 0.5 m$