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levacccp
3 months ago
11

A tube with a cap on one end, but open at the other end, produces a standing wave whose fundamental frequency is 130.8 Hz. The s

peed of sound is 343 m/s. a. If the cap is removed , what is the new fundamental frequency?
Physics
1 answer:
Softa [3K]3 months ago
8 0

Answer:

A. 261.6 Hz

B. 0.656 m

Explanation:

A.

When a tube is open on one side and closed on the other, the fundamental frequency is given by:

F1 = V/4*L

Where,

F1 = fundamental frequency

V = speed of sound

L = length of the tube

If the tube is open at both ends:

F'1 = V/2*L

Where:

F'1 = new fundamental frequency

Thus,

F'1 = 2 * F1

= 2 * 130.8

= 261.6 Hz

B.

F1 = V/4*L

Or

F'1 = V/2*L

Given:

V = 343 m/s

F1 = 130.8

L = 343/(4 * 130.8)

= 0.656 m

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
inna [3103]

Answer:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + z^{2} k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + v^{2}k

The normal vector N is computed as

N = S_{u}XS_{v}

Where:

S_{u} = =

S_{v} = =

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, v^{2}>. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + uv^{2}

Thus

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu ≈ 3077.34

8 0
2 months ago
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