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PART ONE
An open vertical tube has water in it. A
tuning fork vibrates over its mouth. As the
water level is lowered in the tube, the eighth resonance is heard when the water level is
217.5 cm below the top of the tube.
(IMAGE)
What is the wavelength of the sound wave?
The speed of sound in air is 343 m/s.
Answer in units of cm.

PART TWO
What is the frequency of the sound wave; i.e.,
the tuning fork?
Answer in units of s^−1

PART THREE
The water continues to leak out the bottom
of the tube.
When the tube next resonates with the tuning fork, what is the length of the air column?
Answer in units of cm.

PART ONE An open vertical tube has water in it A tuning fork vibrates over its mouth As the water level is lowered in the tube the eighth resonance is heard whe class=

Respuesta :

opudodennis

Answer:

(a) 58 cm

(b) 591.4 Hz

(c) 246.5 cm

Explanation:

(a)

For closed organ pipe, the wavelength is given by

[tex]\lambda=\frac {4L}{2n-1}[/tex] where [tex]\lambda[/tex] is wavelength, L is the length of the pipe and n is the resonance

Substituting 217.5 m for L and 8 for n then the wavelength of the sound will be

[tex]\lambda=\frac {4\times 217.5}{(2\times 8)-1}=58 cm[/tex]

(b)

We know that the speed of sound, v is a product of frequency and wavelength.

[tex]v=\lambda f[/tex]

Making frequency the subject of the formula then

[tex]f=\frac {v}{\lambda}[/tex] and since the velocity is given as 343 m/s and wavelength is already found in part a above but now we change units from cm to m then

[tex]f=\frac {343}{0.58}=591.3793103\\\boxed {\approx 591.4 Hz }[/tex]

(c)

We know that

[tex]\lambda=\frac {4L}{2n-1}[/tex]

The next resonance will be the 9th and we know the wavelength hence making L the subject then

[tex]L=\frac {\lambda \times (2n-1)}{4}\\\frac {58\times (2*9-1)}{4}\\\=\boxed{ 246.5 cm}[/tex]

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