Question

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.

Short answer

1. The displacement of the string particle at x = 0.20 m is 0.04 m .
2. The displacement of the string particle at x = 0.30 m is 0 .
3. The transverse velocity of the string particle at t = 0.50 s is 0 .
4. The transverse velocity of the string particle at t = 1.0 s is -0.13 m/s.
5. The graph is drawn below.

Step-by-step solution

Given data

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$t=0.50s,x_1=0,x_2=0.40\mathrm{m}$

Understanding the concept of resonant frequency 

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, $u\left(x,t\right)=\frac{dy}{dt}.......\left(5\right)$

Step 3(a): Calculation of displacement of string particle at x = 0.2 m

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

Step 4(b): Calculation of displacement of string particle at x = 0.3 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

Step 5(c): Calculation of transverse velocity at t = 0.50 s

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

Step 6(d): Calculation of transverse velocity at t = 1.0 s

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

Step 7(e): Plotting the graph of the wave

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: