91Ó°ÊÓ

The given wave equation is, $D\left(x,t\right)=\left(3.0cm\right)×\sin \left[2\mathrm{π}\left(\raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$2.4\mathrm{m}$}\right.\right)+t/\left(0.20s\right)+1\right]$that can be compared with the general wave equation as,

role="math" localid="1649783209651" $f\left(x,t\right)=A\sin \left(kx+\mathrm{Ó\neg }t+\mathrm{φ}\right)$

We have to determine the direction in which the wave is moving.

Considering the component of wave inside the 'sin' $\left[2\mathrm{Ï€}\left(\raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$2.4\mathrm{m}$}\right.\right)+t/\left(0.20s\right)+1\right]$that should be constant, we can say, if the time increases the coefficient of x should be decrease. Thus, the wave must be move along the negative x-axis.

Comparing the given wave equation with the general wave equation, $f\left(x,t\right)=A\sin \left(kx+\mathrm{Ó\neg }t+\mathrm{φ}\right)$, we can conclude that the angular wavelength role="math" localid="1649785043838" $k=\frac{2\mathrm{Ï€}}{2.4m}$and angular speed role="math" localid="1649784692609" $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{0.20s}$.

We can determine the wave speed, the frequency, and the wave number using the above relations.

The frequency of the wave,

$$\begin{gathered} f=\frac{\mathrm{Ó\neg }}{2\mathrm{Ï€}} \\ f=\frac{2\mathrm{Ï€}}{\left(0.20s\right)×2\mathrm{Ï€}} \\ f=5Hz \end{gathered}$$

Now, the angular wavelength or wave number is,

$$\begin{gathered} k=\frac{2\mathrm{π}}{\mathrm{λ}} \\ k=\frac{2\mathrm{π}}{2.4m} \\ k=2.6rad/m \end{gathered}$$

Now, the wavelength of the wave, $\mathrm{λ}=2.4m$

The wave speed can be obtained as, $v=f\mathrm{λ}$

Therefore,

$$\begin{gathered} v=\left(5hz\right)×\left(2.4m\right) \\ v=12m/s \end{gathered}$$

We have to find the displacement of the string at time t = 0.50 s, and position of wave at x = 0.20 m.

Substitution the given value of time and position in the wave equation, we can get the magnitude of displacement of the string as,

$$\begin{gathered} D\left(x,t\right)=\left(3.0cm\right)×\sin \left[2\mathrm{π}\left\{\left(\frac{0.20}{2.4}\right)+\left(\frac{0.50}{0.2}\right)+1\right\}\right] \\ D\left(x,t\right)=\left(3.0cm\right)×\sin 2\mathrm{π}\left(3.58\right) \\ \mathrm{D}\left(\mathrm{x},\mathrm{t}\right)=\left(3.0\mathrm{cm}\right)×0.38 \\ \mathrm{D}\left(\mathrm{x},\mathrm{t}\right)=1.15\mathrm{cm} \end{gathered}$$

As the wave is moving along the negative x-axis, the displacement of wave should be$-1.15cm.$