Atransverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 86 m/s. at t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. the maximum transverse speed of the string particle at x = 0 is 18 m/s. (a) what is the frequency of the wave? (b) what is the wavelength of the wave? if the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?

a) 71.62 Hz

b) 1.2 m

c) 0.04m

d) 5.23 1/m

e) 450 1/s

f)  Ф=π/2

g) minus sign

Explanation:

Hi!

Let's start with the wave equation and then write all the conditions that aregiven to us:

The text says that at t=0 and x=0  the transverse displacement is 4.0 cm:

-- (1)

And also it says that the particle is not moving this means that its transversal velocity is zero:

For this to be zero, and not reduce to a trivial solution (y=0)

cos(Ф) = 0     --->  Ф=π/2 -- (2)

Repacing (2) in (1) we get:

Therefore

[text]y_{m}=0.04[/text] --- (3)

Now it says that the maximum transverse speed at x=0 is 18m/s this means that:

ωA=18m/s

and since A=0.04m we get that

ω = 450 1/s --(4)

The relation between the frequency (f) and angular frequency(ω) is given by:

ω=2πf

Therefore:

a)

The frequency of the wave is:

    f = ω/2π = 71.62 Hz

b)

To calculate the wavelength we must use the following expression between the wavelength, the frequency and the velocity of the wave:

v=fλ

so the wavelength is:

   λ=v/f =1.2 m

c)

We already have calculated this value in eq (3)

   ym=0.04m

d)

The relationship between the wavenumber and the wavelength is:

k=2π/λ

Therefore

  k=5.23 1/m

e)

We also calculated this one before at eq(4)  

   ω = 450 1/s

f)

From eq(2)

     Ф=π/2

g)

The correct choice of the sign of ω is the minus sign since the wave is traveling to the positive direction of the x axis

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3. dark matter

empty space could also be a simple vacuum