Answer:
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:
[tex]0.04=y(0,0)=y_{m}sin(\phi)[/tex] -- (1)
And also it says that the particle is not moving this means that its transversal velocity is zero:
[tex]0m/s =\frac{dy}{dt}(0,0)=(+-)\omega y_{m}cos(\phi)[/tex]
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:
[tex]0.04=y(0,0)=y_{m}sin(\phi) = y_{m}sin(\pi /2)=y_{m}[/tex]
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
