Question

Take a derivative of the equation for y with respect to t. The variable x is to be taken as constant. Use the chain rule in which you take a derivative first of the function and then of the argument.

y = y_m sin(kx - ωt + φ).

Answer 1

y = ym*sin (kx - wt + )

Derivative of the function will be

d(y)/dt = d[ym*sin (kx - wt + )]/dt

= sin (kx - wt + )*d(ym)/dt + ym*d(sin (kx - wt + ))/dt

Now we know that

d(sin At)/dt = A*cos At

d(C)/dt = 0

where A and C are constant

So,

d(y)/dt = 0 + ym*(-w)*cos (kx - wt + )

dy/dt = -w*ym*cos (kx - wt + )