A sinusoidal wave of frequency 50 Hz travels along a string at velocity of 28 m/s. At a given instant the displacement and velocity of a certain point in the string are 24 mm and 1.2 m/s respectively. Taking the certain point and given instant to be x=0, t=0, derive the traveling wave equation which gives the displacement of any point on the string as a function of position x, and time t.
A point in the string has an acceleration of 1800 m/s², at a time 3.0 ms before the instant specified above. What is the minimum distance possible between this and the point x=0
λ=v/f , k=2π/λ , y(x,t)= Asin(kx-ωt+[itex]\varphi[/itex])
I assume ∂²x/∂t²= -Aω²Sin(kx-ωt+[itex]\varphi[/itex])
The Attempt at a Solution
I would really appreciate the second part explained to me.
I got the first part:
λ= v/f = 28/50 = 0.56 m and k= 2π/λ = 2π/0.56 = 11.2 rad/m and ω= 2πf = 100π
the velocity of the displaced point is obtained using a differential equation:
dy/dt = -Aωcos(kx-ωt+[itex]\varphi[/itex])
y= 0.024 = Asin([itex]\varphi[/itex]) at x=0, t=0
dy/dy= 1.2= -Acos([itex]\varphi[/itex])
solving the simultaneous equations we have [itex]\varphi[/itex] = -1.41 rad and A = -0.024 m
∴Equation of travelling wave is y= -0.024sin(11.2x-314t-1.41).
no idea about the next part. Can someone do it for me?