For a particular transverse standing wave on a long string, one of an antinodes is at x = 0and an adjacent node is at x = 0.10 m. The displacement y(t)of the string particle at x = 0is shown in Fig.16-40, where the scale of y theaxis is set by ${\mathbf{y}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{cm}$. When t = 0.50 s, What is the displacement of the string particle at (a) x = 0.20 mand x = 0.30 m (b) x = 0.30 m? What is the transverse velocity of the string particle at x = 0.20 mat (c) t = 0.50 sand (d) t = 0.1 s ? (e) Sketch the standing wave atfor the range x = 0to x = 0.40 m.

String displacement$y_s=4.0\mathrm{cm}$

Initial time $t=0.50s$

To find the displacement we are given ${\mathrm{t}}_1=0.50\mathrm{s},{\mathrm{x}}_1=0.20\mathrm{m},{\mathrm{x}}_2=0.30\mathrm{m}$

To find the transverse velocity we are given $x=0.20\mathrm{m},{\mathrm{t}}_1=0.50\mathrm{s},{\mathrm{t}}_2=1.0\mathrm{s}$

To sketch the standing wave we are given$t=0.50s,x_1=0,x_2=0.40\mathrm{m}$

We can write the standing wave equation and find the positions by substituting the given data. To find the transverse velocity we can use the given formula.

Formula:

The equation of the standing wave,$y\left(x,t\right)=\left[y_s\sin \omega t\right]\cos kx.........\left(1\right)$

Angular frequency of the wave, $\omega =\frac{2\pi }{T}.......\left(2\right)$

Wave vector of the wave, $k=\frac{2\pi }{\lambda }.......\left(3\right)$

Wavelength, $\mathrm{\lambda }=4\times \mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{two}\mathrm{successive}\mathrm{antinodes}\mathrm{and}\mathrm{node}......\left(4\right)$

The transverse velocity of the string particle, $u\left(x,t\right)=\frac{dy}{dt}.......\left(5\right)$

From the Figure 16-40, we can see that time period is 2.0 s .

Therefore using equation (2) $\mathrm{\omega }=\mathrm{\pi }\mathrm{rad}/\mathrm{s}$,

It is given that anti node is at x = 0 and node is at x = 0.10 m . This implies that the equation of the standing wave is given as:

$$\begin{gathered} kx=\frac{\pi }{2} \\ k\left(0.10\right)=\frac{\pi }{2} \\ k=5\pi \mathrm{rad}/\mathrm{m} \end{gathered}$$.

Using equation (1), the displacement of the wave at x = 0.20 m is given as:

$$\begin{gathered} y=\left[-0.04\right]\sin \left(\pi \times 0.50\right)\cos \left(5\pi \times 0.20\right) \\ =0.04\mathrm{m} \end{gathered}$$

Hence, the value of displacement is 0.04 m

Using equation (1), it is given that the displacement of the particle at x = 0.30 m

$$\begin{gathered} y=\left[0.04\right]\sin \left(\pi \times 0.50\right)\cos \left(5\pi \times 0.30\right) \\ =0 \end{gathered}$$

Hence, the value of displacement is 0

Transverse velocity of the particle at t = 0.50 s :

Using equation (v), we obtain the transverse velocity at t = 0.50 s and x = 0.20 m .

$$\begin{gathered} u\left(x,t\right)=\frac{d}{dt}\left[y_s\cos kx\right]\sin \omega t \\ =-\cos \omega t\times \left[y_s\cos kx\right] \\ =-\left[\left(\cos \left(\pi \times 0.50\right)\right)\right]\pi \times \left[0.04\cos \left(5\pi \times 0.20\right)\right] \\ =0 \end{gathered}$$

Hence, the value of transverse velocity is 0

Similarly using equation (v), the transverse velocity of the particle at t = 1.0 s:

$$\begin{gathered} u\left(x,t\right)=-\left[\left(\cos \left(\pi \times 0.1\right)\right)\right]\pi \times \left[0.04\cos \left(5\pi \times 0.20\right)\right] \\ =-0.13\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of transverse velocity is -0.13 m/s

Fig: The sketch of the standing wave at t = 0.50 s when $0\le x\le 0.40\mathrm{m}$can be plotted as: