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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.