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2 years ago

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A column of argon is open at one end and closed at the other. The shortest length of such a column that will resonate with a 200

Hz tuning fork is 42.5 cm. The speed of sound in argon must be: Group of answer choices

1 answer:

ziro4ka [17]2 years ago

7 0

Answer:

The speed of sound in the argon is 340 m/s.

Explanation:

Given;

fundamental frequency, f₀ = 200 Hz

length of the pipe, L = 42.5 cm = 0.425 m

A pipe that is open at one end and closed at another end is known as a closed pipe.

The wavelength for the first harmonic is calculated as;

L = Node -------> Antinode

L = λ/4

λ = 4L

The speed of the sound is calculated as;

v = fλ

where;

v is the speed of the sound

v = f x 4L

v = 200 x (4 x 0.425)

v = 340 m/s

Therefore, the speed of sound in the argon is 340 m/s.

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a

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b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

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The centripetal force acting on the bob is mathematically represented as

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Now From the diagram we see that this force is equivalent to

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So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

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So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

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=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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