A string oscillates according to the equation$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{cm}\mathbf{)}\mathbf{sin}\mathbf{\left[}\left(\frac{\pi }{3}{\mathrm{cm}}^{-1}\right)\mathbf{x}\mathbf{\right]}\mathbf{cos}\mathbf{\left[}\left(40{\mathrm{\pi s}}^{-1}\right)\mathbf{t}\mathbf{\right]}$What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position$\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$when$\mathit{t}\mathbf{=}\frac{\mathbf{9}}{\mathbf{8}}\mathit{s}$?

The string oscillates according to the equation,

$\mathrm{y}\text{'}=(0.50\mathrm{cm})\sin \left[\left(\frac{\mathrm{\pi }}{3}{\mathrm{cm}}^{-1}\right)\mathrm{x}\right]\cos \left[\left(40{\mathrm{\pi s}}^{-1}\right)\mathrm{t}\right]$

We can find the amplitude, wave number, and angular speed of the two waves whose superposition gives the given wave by comparing the given equation of the wave with the general equation of the standing wave. From it, we can find the speed of the two waves, and the distance between the nodes using the corresponding relations. The transverse speed of a particle of the string at the given position and time can be calculated by taking the derivative of the displacement of the wave.

Formulae:

The wavelength of an oscillation,$\lambda =\frac{2\mathrm{\pi }}{k}.......\left(1\right)$

The wavelength of a standing wave,$\lambda =\frac{2L}{n}.......\left(2\right)$

The angular frequency of the wave,$\omega =2\pi f.......\left(3\right)$

The velocity of the wave,$v=\lambda f.......\left(4\right)$

The velocity of the wave,$v=\sqrt{\frac{T}{\mu }}......\left(5\right)$

The time period of the standing wave, $t=\frac{2L}{nv}......\left(6\right)$

The given wave equation is

$\mathrm{y}\text{'}=(0.50\mathrm{cm})\sin \left[\left(\frac{\mathrm{\pi }}{3}{\mathrm{cm}}^{-1}\right)\mathrm{x}\right]\cos \left[\left(40{\mathrm{\pi s}}^{-1}\right)\mathrm{t}\right]$

General equation of the standing wave of two waves is given as:

$y\text{'}\left(\left\{2y_m\sin \left(kx\right)\right\}\cos \omega t\right)$

Comparing the above two wave equations, we get:

$$\begin{gathered} 2y_m=0.50\mathrm{cm} \\ {y}_m=0.25\mathrm{cm} \end{gathered}$$

Therefore, amplitude of the two waves whose superposition gives the given wave is 0.25 cm .

Wave number of the combining waves is the same as that of the standing wave.

Comparing the given equation with the general equation of the standing wave, we get the wavenumber as given:

$k=\frac{\mathrm{\pi }}{3}{\mathrm{cm}}^{-1}$

Using equation (1) and the given values, we get the wavelength as given:

$$\begin{gathered} \lambda =\frac{2\mathrm{\pi }}{\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)} \\ =6\mathrm{cm} \end{gathered}$$

Again, comparing the given wave equation with the general equation of wave we get the angular frequency as given:

role="math" localid="1661164967750" $\omega =40\mathrm{\pi }$

Again, using equation (3) and the given value, we get the frequency of the wave as given:

$$\begin{gathered} f=\frac{\mathrm{\omega }}{2\mathrm{\pi }} \\ =\frac{40\mathrm{\pi }}{2\mathrm{\pi }} \\ =20\mathrm{Hz} \end{gathered}$$

Speed of the waves using equation (4), we get

$$\begin{gathered} v=20\left(6\right) \\ =120 \\ ~1.2\times {10}^2\mathrm{cm}/\mathrm{s} \end{gathered}$$

Therefore, the speed of the two waves whose superposition gives the given wave is $1.2\times {10}^2\mathrm{cm}/\mathrm{s}$.

The distance between nodes is given by:

$$\begin{gathered} d=\frac{\mathrm{\lambda }}{2}\left(\mathrm{from}\mathrm{equation}\left(2\right)\mathrm{for}\mathrm{n}=1\right) \\ =\frac{6}{2} \\ =3\mathrm{cm} \end{gathered}$$

Therefore, the distance between nodes is$3\mathrm{cm}$

Displacement of the particles on the string of two waves using superposition principle is given by:

$y\text{'}=\left\{2y_m\sin \left(kx\right)\right\}\cos \omega t)$

Hence, speed of the particles on the string is

$$\begin{gathered} v=\frac{dy\text{'}}{dt} \\ =\frac{dy}{dt}\left[\left\{2y_m\sin \left(kx\right)\right\}\cos \omega t)\right] \\ =-\omega 2y_m\sin \left(kx\right)\sin \left(\omega t\right) \\ =-\left(40\mathrm{\pi }\right)\left(0.50cm\right)\sin \left(\frac{\mathrm{\pi }}{3}x\right)\sin \left(40\mathrm{\pi t}\right) \end{gathered}$$