Question

A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f = 770 Hz. For what values of L will the tube resonate with the speaker? (Assume that the speed of sound in air is 343 m/s.) m (lowest possible value) m (second lowest possible value) m (third lowest possible value)

Answer 1

Answer:

- The lowest possible value of the length L1 = 0.45/2

L1 = 0.225m

- Second lowest possible value

L2 = 0.45m

- Third lowest possible value

L3 = 3(0.45)/2

L3 = 0.675m

Explanation:

Before calculating the length of the resonance tube, we need to know the wavelength produced by the wave.

Using the expression

v = Foλ

V is the velocity of wave

Fo is the resonance frequency

λ is the wavelength

From the formula

λ = v/Fo

λ = 343/770

λ = 0.45m

For an open pipe, the first resonant length L1 = λ/2

Second resonant length L2= λ

Third resonant length L3 = 3λ/2

The lowest possible value of the length L1 = 0.45/2

L1 = 0.225m

Second lowest possible value

L2 = 0.45m

Third lowest possible value

L3 = 3(0.45)/2

L3 = 0.675m