Question

Two travelling waves produce a standing wave represented by equation $y=1.0mm\cos \left(1.57c{m}^{-1}\right)x\sin \left(78.5s^{-1}\right)t$, the node closest to the origin in the region $x>0$ will be at $x=\_\_\_\_\_\_\_\_cm$.

Answer 1

Step 1: Given

Equation of wave: $y=A\cos kx\sin \omega t=1.0mm\cos \left(1.57c{m}^{-1}\right)x\sin \left(78.5s^{-1}\right)t$

Amplitude: $A=1mm$

Angular wave number: $k=1.57c{m}^{-1}$

Angular frequency: $\omega =78.5s^{-1}$

Step 2: Formula Used

At node $y=0$, $\cos kx=0$

Step 3: Find the value of x

Find an expression for x using the formula

$$\begin{gathered} \cos kx=0 \\ \Rightarrow \cos kx=\cos \frac{\pi }{2} \\ \Rightarrow kx=\frac{\pi }{2} \\ \Rightarrow x=\frac{\pi }{2k} \end{gathered}$$

Calculate the value of x by substituting the values

$$\begin{gathered} x=\frac{\pi }{2k} \\ =\frac{3.14}{2\times 1.57} \\ =\frac{3.14}{3.14} \\ =1cm \end{gathered}$$

Hence, the closest node will be at $1cm$.