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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 f1 = v/λ1= v/4L  and the higher harmonic frequencies are odd multiples of the fundamental frequency. Generally the nth harmonic is given by,
fn = (2n – 1)f1.
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, f2 = 3f1= 3 x 553.33 = 1659.99 Hz.
And the third harmonic is f3 = 5f1 = 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; fn= (2n-1)f1.

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