Respuesta :
Given,
[tex]y=4.25\times10^{-3}\sin \lbrack200\pi t-(\frac{2\pi x}{800})\rbrack[/tex]The wave equation of a particle is,
[tex]y=A\sin (\omega t-kx)[/tex]Where
• A is the amplitude.
,• ω is the angular velocity.
,• k is the wavenumber.
On comparing the two equations,
Amplitude, A=4.25×10⁻³
Angular velocity, ω=200π rad/s200
The wavenumber, k=2π/800 /m
The velocity of the particle is equal to the time derivative of its displacement.
Thus, the velocity is,
[tex]\begin{gathered} v=\frac{dy}{dt} \\ =\frac{d}{dt}(A\sin (\omega t-kx))_{} \\ =A\omega\cos (\omega t-kx) \end{gathered}[/tex]On substituting the known values, the velocity is,
[tex]\begin{gathered} v=4.25\times10^{-3}\times200\pi\times\cos \lbrack200\pi t-(\frac{2\pi x}{800})\rbrack \\ =0.85\pi\cos \lbrack200\pi t-(\frac{2\pi x}{800})\rbrack \end{gathered}[/tex]The wavelength is given by,
[tex]\lambda=\frac{2\pi}{k}[/tex]On substituting the known values,
[tex]\begin{gathered} \lambda=\frac{2\pi}{(\frac{2\pi}{800})} \\ =800\text{ m} \end{gathered}[/tex]Thus the wavelength is 800 m
The frequency is given by,
[tex]f=\frac{2\pi}{\omega}[/tex]On substituting the known values,
[tex]\begin{gathered} f=\frac{2\pi}{200\pi} \\ =0.01\text{ Hz} \end{gathered}[/tex]Thus the frequency is 0.01 Hz
The period is given by,
[tex]T=\frac{1}{f}[/tex]On substituting the known values,
[tex]\begin{gathered} T=\frac{1}{0.01} \\ =100\text{ s} \end{gathered}[/tex]Thus the time period of the wave is 100 s
