91Ó°ÊÓ

Transverse wave$y_1=0.050cos(\mathrm{Ï€³\ae }+4\mathrm{Ï€³Ù})$

Transverse wave$y_2=0.050cos(\mathrm{Ï€³\ae }+4\mathrm{Ï€³Ù})$

We use the concept of standing waves. We calculate the sum of the given waves, and then, we find the smallest value of x. We can find the derivative of the resultant equation for time to get velocity. Using that velocity, we can find the time where velocity is zero.

Formula:

$\cos \mathrm{Î\pm }+\cos \mathrm{β}=2\cos \left(\frac{\mathrm{Î\pm }+\mathrm{β}}{2}\right)\cos \left(\frac{\mathrm{Î\pm }-\mathrm{β}}{2}\right).........\left(1\right)$

The resultant of the standing waves is the addition of two waves, therefore

$$\begin{gathered} y=y_1+y_2 \\ =0.050\cos \left(\mathrm{Ï€}x-4\mathrm{Ï€}t\right)+0.050\cos \left(\mathrm{Ï€}x+4\mathrm{Ï€}t\right) \\ =0.050\left[\cos \left(\mathrm{Ï€}x-4\mathrm{Ï€}t\right)+\left(\mathrm{Ï€}x+4\mathrm{Ï€}t\right)\right] \end{gathered}$$

Applying the formula of equation (i), we get the resultant wave as:

$$\begin{gathered} y=0.10\cos \left(\frac{\left(\mathrm{Ï€}x-4\mathrm{Ï€}t+\mathrm{Ï€}x+4\mathrm{Ï€}t\right)}{2}\right)\cos \left(\frac{\left(\mathrm{Ï€}x-4\mathrm{Ï€}t+\mathrm{Ï€}x+4\mathrm{Ï€}t\right)}{2}\right) \\ =0.10\cos \left(\frac{2\mathrm{Ï€}x}{2}\right)\cos \left(-\frac{8\mathrm{Ï€}t}{2}\right) \\ =0.10\cos \left(\mathrm{Ï€}x\right)\cos \left(4\mathrm{Ï€}t\right).......\left(2\right) \end{gathered}$$

For the smallest value of x we can write,

We know cosine angle has value 0 at $\frac{\mathrm{Ï€}}{2}$, so we get,
$$\begin{gathered} \mathrm{Ï€}x=\frac{\mathrm{Ï€}}{2} \\ x=\frac{1}{2} \\ =0.05m \end{gathered}$$

At the smallest value of 0.05 m, it corresponds to a node.

The value of the first time the particle at x=0 has zero velocity:

We find the derivative of the resultant wave, y with respect to t, and we get velocity of the wave as:

$$\begin{gathered} v=\frac{dy}{dt} \\ =\frac{d\left(0.10\cos \left(\mathrm{Ï€}x\right)\cos \left(4\mathrm{Ï€}t\right)\right)}{dt}(fromequation(2\left)\right) \\ =d0.10\cos \left(\mathrm{Ï€}x\right)\left[-4\mathrm{Ï€}\left(\sin \left(4\mathrm{Ï€}t\right)\right)\right] \end{gathered}$$

From the above velocity equation, we know that the value of $\sin \left(4\mathrm{Ï€}t\right)=0$ when the values of t will be
$t=0,\frac{1}{4},\frac{1}{2},\frac{3}{4},1,…...$

So we get the first time particle at x=0 with velocity zero at 0 s.

The value of the second time the particle at x=0 has zero velocity:

From part (b), we can see that the second time the particle is at x=0 it has zero velocity at the given time:

$\begin{array}{l}t=\frac{1}{4}\\ =0.25s\end{array}$

Hence, the required value of second time is 0.25 s

The value of the third time the particle is at x=0 has zero velocity:

Again, from part (b) we can see that the third time the particle is at x=0 it has zero velocity at:

$\begin{array}{l}t=\frac{1}{2}\\ =0.50s\end{array}$

Hence, the required value of time is 0.50 s