Respuesta :
Answer:
(a)
[tex]f=20Hz\\And \\w=125.7s^{-1}[/tex]
(b)
[tex]K=3.490m^{-1}[/tex]
Explanation:
We know that the speed of any periodic wave is given by:
v=f×λ
The wave number k is given by:
K=2π/λ
Given data
Amplitude A=2.50mm
Wavelength λ=1.80m
Speed v=36 m/s
For Part (a)
For the wave frequency we plug our values for v and λ.So we get:
v=f×λ
[tex]36.0m/s=(1.80m)f\\f=\frac{36.0m/s}{1.80m}\\ f=20Hz\\[/tex]
And the angular speed we plug our value for f so we get:
[tex]w=2\pi f\\w=2\pi (20)\\w=125.7s^{-1}[/tex]
For Part (b)
For wave number we plug the value for λ.So we get
K=2π/λ
[tex]K=\frac{2\pi }{1.80m}\\ K=3.490m^{-1}[/tex]
The answers for this question are
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Given to us:
Amplitude A= 2.50 mm,
Wavelength λ= 1.80 m,
Velocity V= 36.0 m/sec,
a.) To find out frequency and angular frequency of the wave,we know
[tex]frequency=\dfrac{Velocity}{Wavelength}[/tex]
[tex]f=\dfrac{V}{\lambda}[/tex]
Putting the numerical value,
[tex]f=\dfrac{36}{1.80}\\\\f= 20\ Hz[/tex]
Therefore, the frequency of this wave is 20 Hz.
Also, for angular frequency
[tex]f=2\pi \omega[/tex]
Putting the numerical value,
[tex]\omega=2\pi f \\\\\omega=2\times\pi \times 20\\\\\omega= 125.71428\ s^{-1}[/tex]
Therefore, the angular frequency of this wave is [tex]125.71428\ s^{-1}[/tex].
b.) To find out the wave number of the wave (k),
The wave number for an EM field is equal to 2π divided by the wavelength(λ) in meters.
[tex]k=\dfrac{2\pi}{\lambda}[/tex]
Putting the numerical value,
[tex]k=\dfrac{2\pi}{\lambda}\\\\k=\dfrac{2\pi}{1.80}\\\\k=3.4920\ m^{-1}[/tex]
Therefore, the wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
Hence, for this wave
a.) frequency of the wave is 20 Hz.
b.) angular frequency of the wave is [tex]125.71428\ s^{-1}[/tex]
c.) wave number of the wave is [tex]k=3.4920\ m^{-1}[/tex].
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https://brainly.com/question/3813804
