Answer:
The equation of displacement is [tex]y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})[/tex].
Explanation:
Given that,
Distance = 2.50 m
We need to calculate the equation of wave
Using general equation of wave
[tex]y=A\sin(\omega t-kx+\phi)[/tex]....(I)
Where, A = amplitude
t = time
x = displacement
[tex]\phi[/tex] = phase difference
Put the value in the equation
At t = 0, x = 0, y =A
[tex]A=A\sin(0+\phi)[/tex]
[tex]\sin\phi=1[/tex]
[tex]\phi=\dfrac{\pi}{2}[/tex]
From equation (I)
[tex]y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})[/tex]
Hence, The equation of displacement is [tex]y=A\sin(\omega t-2.50 k+\dfrac{\pi}{2})[/tex].
The equation of displacement in terms t will be given as
[tex]y=ASin(wt-2.5k+\dfrac{\pi}{2} )[/tex]
The oscillation is defined as the regular movement of the wave from its mean position.
We have the following data given in the question.
Distance = 2.50 m
For finding the displacement of the wave
Using the general equation of wave
[tex]y=ASin(wt-kx+\phi)[/tex]
Where,
A = amplitude of the wave
t = time period of the wave
x = displacement
[tex]\phi[/tex] = phase difference
Put the value in the equation
At t = 0, x = 0, y =A
[tex]A=ASin(0+\phi)[/tex]
The equation will become
[tex]Sin\phi=1[/tex]
[tex]\phi=\dfrac{\pi}{2}[/tex]
Put the value of [tex]\phi[/tex] in equation 1
[tex]y=ASin(wt-2.5k+\dfrac{\pi}{2})[/tex]
Thus the equation of displacement will be given as
[tex]y=ASin(wt-2.5k+\dfrac{\pi}{2} )[/tex]
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