91Ó°ÊÓ

Figure for the resultant wave is given.

An antinode A travels from extreme upward to extreme downward in time, t=6.0 ms

Tick marks along the axis are separated by,x=10 cm

Height is, H=1.80 cm

One of the superimposed wave is of the form,$y(x,t)=y_{m}sin(kx+\mathrm{Ó\neg }t)$

We can find the valueof the amplitude of the standing wave by comparing the given equation with the general equation for the standing wave. The wavelength and period of the standing wave can be predicted from the given figure. The frequency of the wave can be calculated from the period using the corresponding formula. This can be used to find the angular speed. From the equation of the other wave, we can find the sign of angular speed.

Formulae:

The wavenumber of the wave,$k=\frac{2\mathrm{π}}{\mathrm{λ}}.......\left(1\right)$

The frequency of the wave,$f=\frac{1}{T}.....\left(2\right)$

The angular frequency of the wave,$\mathrm{Ó\neg }=2\mathrm{Ï€´Ú}…..\left(3\right)$

To form a standing wave, the equations of waves should be

$y\left(x,t\right)=y_m\sin \left(kx+\mathrm{Ó\neg }t\right).....\left(4\right)$

$y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right).....\left(5\right)$

Therefore, according to the superposition principle, the equation of the resultant wave is

$y^{\text{'}}=y_{m}sin(kx+\mathrm{Ó\neg }t)+y_{m}sin(kx-\mathrm{Ó\neg }t)$

$y^{\text{'}}=\left(2y_{m}sin\left(kx\right)cos\mathrm{Ó\neg }t\right)$

From the given figure we can write that the amplitude of the standing wave of the two waves is given as:

$$\begin{gathered} y_m^{\text{'}}=\frac{H}{2} \\ =\frac{1.80}{2} \\ =0.9\mathrm{cm} \\ =9.0\mathrm{mm} \end{gathered}$$

The amplitude of one of the waves that superimposes to form the given standing wave is given as:

$$\begin{gathered} y_m=\frac{y_m^{\text{'}}}{2} \\ =\frac{9}{2} \\ =4.5mm \end{gathered}$$

Therefore, the value of$y_m$ for the other wave is 4.5 mm

From the given figure, we can infer that the wavelength of the standing wave is

$\mathrm{λ}=40\mathrm{cm}$

Using equation (1), we get the wavenumber as:

$$\begin{gathered} \mathrm{k}=\frac{2\left(3.142\right)}{\left(40\right)} \\ =0.1571{\mathrm{cm}}^{-1} \\ =15.71{\mathrm{m}}^{-1} \\ ~16{\mathrm{m}}^{-1} \end{gathered}$$

Therefore, the value of k for the other wave is $6{\mathrm{m}}^{-1}$.

An antinode A travels from extreme upward to extreme downward in time t=6.0 ms

Therefore period of the wave is given as:

$$\begin{gathered} T=12\mathrm{ms} \\ =12×{10}^{-3}\mathrm{s} \end{gathered}$$

Using equation (2) and the time, we get the frequency of the other wave as:

$$\begin{gathered} f=\frac{1}{12×{10}^{-3}} \\ =83.33\mathrm{Hz} \end{gathered}$$

Using equation (3) and the given value of frequency, the angular velocity of the wave is given as:

$$\begin{gathered} \mathrm{Ó\neg }=2\left(3.142\right)\left(83.33\right) \\ =523.46 \\ ~5.2×{10}^2\frac{\mathrm{rad}}{\mathrm{s}} \end{gathered}$$

Therefore, the value of for the other wave is$5.2×{10}^2\frac{rad}{s}$

The other wave equation is

$y(x,t)=y_{m}sin(kx-\mathrm{Ó\neg }t)$

Therefore, the sign of role="math" localid="1661162543996" $\mathrm{Ó\neg }f$or the other wave is negative.