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saveliy_v [14]
3 years ago
10

Is it possible to create a visual model (stop motion) of a wave motion using the differential equation for simple harmonic oscil

lation and how so?
Physics
1 answer:
Alexeev081 [22]3 years ago
8 0

Answer:

  • Yes.

Explanation:

A particular solution for the 1D wave equation has the form

\Psi(x,t) \ = \ A \ sin ( \ k x \ + \omega t \ + \phi \ )

where A its the amplitude, k the wavenumber, ω the angular frequency and φ the phase angle.

Now, for any given position x_0, we can use:

\phi_0 \ = \ \phi \ + \ k x_0

so, the equation its:

\Psi(x_0,t) \ = \ A \ sin ( \omega t \ + \ \phi_0 ).

This is the equation for a simple harmonic oscillation!

So, for any given point, we can use a simple harmonic oscillation as visual model. Now, when we move a \delta distance from the original position, we got:

x_1 = x_0 + \delta

and

\phi_1 = \phi  \ + k x_1

now, this its

\phi_1 = \phi  \ + k ( x_0 + \delta)

\phi_1 = \phi  \ + k  x_0 + k \delta

\phi_1 = \phi_0 + k \delta

So, there its a phase angle difference of k \delta. We can model this simply by starting the simple harmonic oscillation with a different phase angle.

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Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

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What happens when the mass of an object is greter
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The acceleration goes up.
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A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
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Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

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Can someone please help me ASAP
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Answer:

distance = 6.1022 x10^16[m]

Explanation:

To solve this problem we must use the formula of the average speed which relates distance to time, so we have

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distance = x [meters]

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Now we have to convert from light-years to seconds in order to get the distance in meters.

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Now using the formula:

distance = v * time

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