Solve Harmonic Wave Displacement at x=10, t=20s

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To solve for the displacement of a harmonic wave at x = 10 cm and t = 20 s, the wave equation y = A sin(kx - ωt) should be utilized. The amplitude (A) is given as 3.8 cm, and the angular frequency (ω) has been calculated as approximately 5.96. The wave number (k) can be determined using the wavelength, which is 38 cm, leading to k = 2π/λ. Substituting the values into the equation will yield the displacement at the specified coordinates. Proper application of these formulas will provide the required solution for part B of the problem.
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Hello, I am having a problem solving part B of the following problem

A harmonic wave moving in the positive x direction has an amplitude of 3.8 cm, a speed of 36.0 cm/s, and a wavelength of 38.0 cm.

A) Calculate the displacement (in cm) due to the wave at x = 0.0 cm, t = 2.0 s. Assume that the displacement is zero at x = 0 and t = 0.

B) Calculate the displacement due to the wave at x = 10.0 cm, t = 20.0 s.

I got the answer to the first problem by using the equation
y=A x Sin (w x t)

f=velocity / wavelength = (.36m)/(.38m)= 0.94736

Period (T)= 1 / F = 1/0.94736 = 1.055

w= 2 x Pi/T = 2*pi / (1.055) = 5.9556

x= 2.3cm

This was fairly easy because x=0 and t=2 sec.

But I am not sure what to do when x= 10.cm and t= 20sec

Anybody got any tips?
 
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Just use the formula y=asinwt where a=38 cm
 
the formula for an harmonic wave

y = A \sin (kx - \omega t)
 
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