A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by, $$y(x, t) = 0.5 \sin\left(\frac{5\pi}{4}x\right) \cos(200\pi t)$$. What is the speed of the travelling wave moving in the positive $$x$$ direction? ($$x$$ and $$t$$ are in meter and second, respectively)

A standing wave is produced when two identical sinusoidal waves move in opposite directions. For a travelling wave moving in the +$$x$$ direction we may write the displacement as $$y_1=A\sin(kx-\omega t)$$, and for the wave moving in the −$$x$$ direction we write $$y_2=A\sin(kx+\omega t)$$.

Adding the two, we use the trigonometric identity $$\sin P+\sin Q = 2\sin\!\left(\frac{P+Q}{2}\right)\cos\!\left(\frac{P-Q}{2}\right)$$. Substituting $$P=kx-\omega t$$ and $$Q=kx+\omega t$$ we obtain

$$y=y_1+y_2=2A\sin(kx)\cos(\omega t).$$

The given equation of the standing wave is

$$y(x,t)=0.5\sin\!\left(\frac{5\pi}{4}\,x\right)\cos(200\pi t).$$

By comparing this expression with the standard form $$y=2A\sin(kx)\cos(\omega t)$$ we can immediately read the wave parameters:

$$2A=0.5\;\;\Rightarrow\;\;A=0.25\;(\text{m, not needed further}),$$

$$k=\frac{5\pi}{4}\;\text{rad m}^{-1},$$

$$\omega=200\pi\;\text{rad s}^{-1}.$$

The speed $$v$$ of either parent travelling wave is related to $$k$$ and $$\omega$$ by the fundamental wave relation

$$v=\frac{\omega}{k}.$$

Substituting the numerical values, we have

$$v=\frac{200\pi}{\dfrac{5\pi}{4}}.$$

First, notice that the factor $$\pi$$ cancels:

$$v=\frac{200}{\dfrac{5}{4}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal, so

$$v=200 \times \frac{4}{5}.$$

Carrying out the multiplication,

$$v=\frac{800}{5}=160\;\text{m s}^{-1}.$$

Hence, the correct answer is Option C.