**Question 12:** Estimate the frequencies at which a test tube 15 cm long resonates when you blow across its lip.

ANSWER

### Estimation of frequencies in the test tube

Such a test tube will behave like a closed organ pipe when air is blown into it. We know that the lowest or fundamental frequency of a closed organ pipe is given by f_{1} = v/λ_{1}= v/4L and the higher harmonic frequencies are odd multiples of the fundamental frequency. Generally the nth harmonic is given by,

f_{n} = (2n – 1)f_{1}.

Here, length of the tube = 15 cm = 15 x 10^{-2} m

Let speed of the sound = 332 m/s,

Put in the above equation for the frequency of first harmonic.

Putting the values, f_{2} = 3f_{1}= 3 x 553.33 = 1659.99 Hz.

And the third harmonic is f_{3} = 5f_{1} = 5 x 553.33 = 2766.65 Hz.

……….. And so on.

**Note:** You can also find the values of f2 and f3 etc. by applying the above formula; f_{n}= (2n-1)f_{1}.

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