Two waves are generated on a string of length 3.0 , to produce a three-loop standing wave with an amplitude of 1.0 cm. The wave speed is 100 m/s. Let the equation for one of the waves is of the form$\mathit{y}\mathbf{(}\mathit{x}\mathbf{,}\mathbf{}\mathit{t}\mathbf{)}\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{k}\mathit{x}\mathbf{+}\mathit{\omega }\mathit{t}\mathbf{)}$. In the equation for the other wave, What are (a)${\mathit{y}}_{\mathbf{m}}$(b)k, (c) What is$\mathit{\omega }$, and (e) the sign in front of$\mathit{\omega }$?

No. of loops are,n=3

Length of string is,I =3.0 m

Amplitude of standing wave is,$y_m^{\text{'}}=1.0\mathrm{cm}\mathrm{or}0.1\mathrm{m}$$y_m^{\text{'}}=1.0\mathrm{cm}\mathrm{or}0.01\mathrm{m}$

Wave speed of standing wave is,v=100 m/s

Wave equation of one of the wave is,$y(x,t)=y_{m}sin(kx+\omega t)$

We can find the value of ${\mathit{y}}_{\mathbf{m}}$from the amplitude of the standing wave. The wavelength of the standing wave can be calculated from the length of the string and no. of loops using the corresponding relation which is used to find the wave number. The frequency of the wave can be calculated from the wavelength using the corresponding formula. This can be used to find the angular speed.

Formulae:

The expression of the standing wave,$y^{\text{'}}=\left\{2y_m\sin \left(kx\right)\right\}\cos \omega t).....\left(1\right)$

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

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

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

Wavelength of the standing wave,$\lambda =\frac{2l}{n}\cdot \cdot \cdot \cdot \cdot \cdot \left(5\right)$

Using equation (1), we get the value of the amplitude as given:

Given that

$$\begin{gathered} 2{\mathrm{y}}_{\mathrm{m}}=0.01\mathrm{m} \\ {\mathrm{y}}_{\mathrm{m}}=5\times {10}^{-3}\mathrm{m} \\ =0.5\mathrm{cm} \end{gathered}$$

Therefore, the value ofrole="math" localid="1661166430187" ${\mathrm{y}}_{\mathrm{m}}=0.5\mathrm{cm}$

Using equation (5), we get the value of wavenumber as given:

$$\begin{gathered} \lambda =\frac{2\left(3.0\right)}{3}\left(\because \mathrm{the}\mathrm{wave}\mathrm{is}\mathrm{at}\mathrm{third}\mathrm{harmonic}\right) \\ =2.0m \end{gathered}$$

Hence, using equation (2) and the given values, we get the wavenumber as:

$$\begin{gathered} k=\frac{2\left(3.142\right)}{\left(2m\right)} \\ =3.142{\mathrm{m}}^{-1} \\ ~3.1{\mathrm{m}}^{-1} \end{gathered}$$

Therefore, the value of k is $3.1{\mathrm{m}}^{-1}$.

Using equation (4) and the given values, the frequency of the wave as given:

$$\begin{gathered} f=\frac{100}{2} \\ =50\mathrm{Hz} \end{gathered}$$

Angular velocity of the wave using equation (3) is given as:

$$\begin{gathered} \omega =2\left(3.142\right)\left(50\right) \\ =314.2\mathrm{rad}/\mathrm{s} \\ ~314\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the value of$\omega =314\mathrm{rad}/\mathrm{s}$

Wave equation of the other wave is

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

Therefore, the sign of$\omega$ of the other wave is minus.