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mihalych1998 [28]
3 years ago
5

To understand the standard formula for a sinusoidal traveling wave. One formula for a wave with a y displacement (e.g., of a str

ing) traveling in the x direction is y(x,t)=Asin(kx−ωt).
All the questions in this problem refer to this formula and to the wave it describes.
1) What is the phase ϕ(x,t) of the wave? Express the phase in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π
ϕ(x,t)=
2) What is the wavelength λ of the wave? Express the wavelength in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
λ=
3) What is the period T of this wave? Express the period in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
T=
4) What is the speed of propagation v of this wave? Express the speed of propagation in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
v=
Physics
1 answer:
AysviL [449]3 years ago
3 0

Explanation:

The displacement of the string in the x direction is given by :

y(x,t)=A\ sin(kx-\omega t)

Where

A is the amplitude of wave

k is the propagation constant

1. Here, the phase of the wave is (kx-\omega t)

2. The propagation constant of a wave is given by :

k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{k}

3. Since, \omega=2\pi f

f=\dfrac{\omega}{2\pi}

Time period, T=\dfrac{1}{f}

T=\dfrac{2\pi}{\omega}

4. Speed of this wave is given by :

v=f\times \lambda

v=\dfrac{\omega}{2\pi}\times \dfrac{2\pi}{k}

v=\dfrac{\omega}{k}

Hence, this is the required solution.

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PRACTICE ANOTHER A cube of wood having an edge dimension of 19.7 cm and a density of 647 kg/m3 floats on water. (a) What is the
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CHECK COMPLETE QUESTION

The cube of wood having an edge dimension of 19.7cm and a density of 647kg/m3 floats on water

(a) What is the distance from the horizontal top surface of the cube to the water level answer in cm

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Answer:

a)6.29cm

b)2.78 kg

Explanation:

Given:

Let us calculate the volume first, we were given dimension as 19.7cm=0.197m

Volume is (0.197 meters)³ = 0.00764m³

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Given mass is 647 kg/m³ x 0.00774m³ = 4.947kg

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which is equal to the weight of volume of the displaced water. Or the mass, the calculation is the same.

We know that density of fresh water at 20ºC is 998 kg/m³

Then we can calculate the volume of displaced water as

4.947 kg / 998 kg/m³ = 0.00496 m³

We know that the displaced water has a shape of a rectangular solid with 0.197 meters on the two horizontal dimensions, and h as the height to the surface then

V = 0.197²h = 0.00496

0.00496= 0.197²h

h = 0.1278 meters or 12.78 cm

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b) ifthe cube is fully submerged, the volume of the displaced water is 0.00774m³

mass of displaced water is 0.00774m³ x 998 kg/m³ = 7.724 kg

Added mass is the mass of the displaced water – mass of block

= 7.724–4.947 = 2.78 kg

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