Question

Acceleration of point x = 2 m of a transverse wave is 5 j^ m/s² and the slope of waveform is given by $\frac{dy}{dx} = 5 sin \frac{ \pi}{12} x$. Then the speed of the point at that instant is given in m/s as

Answer 1

Transverse progressive wave  y(x, t) = A Sin(ωt - kx)  meters
  Given x = 2 m,  t = t1 

Velocity (in y direction) of a point at x and at time t = t1 is 
    = dy/dt at constant x
    = Aω Cos(ωt - kx)   m/sec

Acceleration (in y direction) of a particle at x at time t = t1 is 
    = d²x/dt² = at constant x = - A ω² Sin(ωt₁ - k x)   m/sec²
 
Slope of waveform at x in a photosnap of the waveform taken at a  particular time t₁
     = dy/dx at Constant t = t₁
     = - A k Cos(ωt₁ - k x)
     =  5 Sin (πx/12) = 5 Cos (π/2 - πx/12)      --- Given

Comparing the two lines:
     ωt₁ = π/2  rad.
     k = π/12  rad/meter 
     A k = 5    =>   A = 60/π  meters

Given acceleration: | - A ω² Sin(ωt₁ - k x) | = 5 m/s²
                             - 60/π * ω² Sin(π/2 - π/12 * 2) = 5 
                             ω² = π / 6√3
                             ω = 0.3023 rad/sec

=>  Velocity at t=t1 and x=2 m,  v = A ω Cos(ω t₁ - k x)   m/s
      v = 60/π * 0.3023 * Cos(π/2 - π/12 * 2)    m/s
         = 2.886 m/s