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dem82 [27]
3 years ago
8

You perform an experiment with a long column of air and a tuning fork. The column of air is defined by a very long vertical plas

tic tube with a circular cross section; the upper end of the tube is open to the outside air and the lower end of the tube is filled with water. The column of air extends from the top of the water to the open end of the tube; the length of the column can be varied by changing the water level in the lower end of the tube. (The columns of air in this problem are longer than those you will have used in lab; this allows us to imagine the use of lower frequency tuning forks than those you will have used in lab.)
ou are using a tuning fork of frequency 256 Hz. For one particular water level, you hear a loud sound when the fork is struck and positioned just above the open end of the tube. When the water level is lowered by 676 mm, you again hear a loud sound.
(a) What is the wavelength of the 256 Hz sound waves in the column of air? Give your answer in meters, not mm. m
(b) What is the speed of sound in the column of air? m/s
Physics
2 answers:
velikii [3]3 years ago
6 0

Answer:

\lambda=4L=1.33m

v=343m/s

Explanation:

We have to take into account the expressions

f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}

if we assume that 256Hz is the fundamental frequency we have

f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m

and for wavelength

\lambda=4L=1.33m

hope this helps!!

horsena [70]3 years ago
5 0

Answer:

a) The wavelength, λ = 1.352 m

b) The speed of the sound in the column air is v = 346.112 m/s

Explanation:

a) Frequency, f = 256 Hz

The water level is lowered by 676 mm. i.e. x = 676 * 10⁻³ m

The length, x, at which the loudest sound will be heard is λ/2

i.e. x =  λ/2

Therefore the wavelength of the 256 Hz sound waves in the column air is given by the relation

λ = 2x

λ = 2 * 676 * 10⁻³

λ = 1.352 m

b) The speed of the sound in the column air:

The speed, v, is given by the relation   v = f  λ

speed, v = 256 * 1.352

v = 346.112 m/s

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Explanation:

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Mass of block = 3 kg

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Coefficient of kinetic friction = 0.20

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Using balance equation of force

N = mg\co\theta

mg\sin\theta-f_{k}=ma

a = g\sin\theta-\mu g\cos30

Put the value into the formula

a=g(\sin30-0.20\cos30)

a=9.8(\dfrac{1}{2}-0.20\times\dfrac{\sqrt{3}}{2})

a=3.20\ m/s^2

The acceleration is 3.20 m/s².

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Using formula of work done

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The final speed of the block is 6.26 m/s.

Hence, This is the required solution.

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