A continuous sinusoidal longitudinal wave is sent along a very long coiled spring from an attached oscillating source. The wave travels in the negative direction of an xaxis; the source frequency is$\mathbf{25}\mathbf{\text{\hspace{0.17em}Hz}}$; at any instant the distance between successive points of maximum expansion in the spring is; the maximum longitudinal displacement of a spring particle is$\mathbf{24}\mathbf{\text{\hspace{0.17em}cm}}$; and the particle at$\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$has zero displacement at time$\mathbf{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$. If the wave is written in the form$\mathbf{s}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{t}\mathbf{)}\mathbf{}\mathbf{=}{\mathbf{s}}_{\mathbf{m}}\mathbf{cos}\mathbf{(}\mathbf{kx}\mathbf{\pm }\mathbf{}\mathbf{\omega t}\mathbf{)}$, what are (a)${\mathbf{s}}_{\mathbf{m}}$, (b)$\mathbf{k}$, (c)$\mathbf{\omega }$, (d) the wave speed, and (e) the correct choice of sign in front of$\mathbf{\omega }$?

1. The distance between two successive points,$\mathrm{\lambda }=24\text{\hspace{0.17em}cm}$
2. At $\mathrm{t}=0\mathrm{s}$,$\mathrm{s}=0.$
3. The maximum longitudinal displacement,${\mathrm{s}}_{\mathrm{m}}=0.30\text{\hspace{0.17em}cm}$
4. Frequency ofthesource,$\mathrm{f}=25\text{\hspace{0.17em}Hz}$
5. The displacement equation,$\mathrm{s}(\mathrm{x},\mathrm{t})={\mathrm{s}}_{\mathrm{m}}\cos (\mathrm{kx}\pm \mathrm{\omega t})$

From the given wave equation, we can find the amplitude using the relation between wavelength and wavenumber. We can use the relation between linear frequency and angular frequency to find the angular frequency. We can use the formula of wave speed to find its value using the given frequency.

Formula:

The angular frequency of a wave,

$\mathbf{\omega }\mathbf{=}\mathbf{2}\mathbf{\pi f}$ …(i)

The wave number of a wave,

role="math" localid="1661509157708" $\mathbf{k}\mathbf{=}\mathbf{2}\mathbf{\pi }\mathbf{/}\mathbf{\lambda }$ …(ii)

The velocity of a wave,

$\mathbf{v}\mathbf{=}\mathbf{\omega }\mathbf{/}\mathbf{k}$ …(iii)

Given that,

$\mathrm{s}(\mathrm{x},\mathrm{t})={\mathrm{s}}_{\mathrm{m}}\cos (\mathrm{kx}\pm \mathrm{\omega t})$

${\mathrm{s}}_{\mathrm{m}}=0.30\text{\hspace{0.17em}cm}$

Therefore, the amplitude${\mathrm{s}}_{\mathrm{m}}$ of the wave is $0.30\text{\hspace{0.17em}cm}$

For wave number using equation (ii), we get

$\begin{array}{c}\mathrm{k}=\frac{2\mathrm{\pi }}{24\text{\hspace{0.17em}cm}}\\ \mathrm{k}=0.26{\text{\hspace{0.17em}cm}}^{\text{-1}}\end{array}$

Therefore, wave number $\mathrm{k}$is$0.26{\text{\hspace{0.17em}cm}}^{\text{-1}}$

For angular frequency using equation (i), we get

$\begin{array}{c}\mathrm{\omega }=2\mathrm{\pi }\left(25\text{\hspace{0.17em}Hz}\right)\\ \mathrm{\omega }=1.6\times {10}^2\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Therefore, angular frequency $\mathrm{\omega }$is$1.6\times {10}^2\text{\hspace{0.17em}rad}/\text{s}$

For wave speed using equation (iii), we get

As $\mathrm{\omega }=1.6\times {10}^2\text{\hspace{0.17em}rad}/\text{s},\mathrm{k}=0.26{\text{\hspace{0.17em}cm}}^{\text{-1}}$

$\begin{array}{c}\mathrm{v}=\frac{(1.6\times {10}^2\text{\hspace{0.17em}rad}/\text{s})}{\left(0.26{\text{\hspace{0.17em}cm}}^{\text{-1}}\right)}\\ \mathrm{v}=6.0\times {10}^2\text{\hspace{0.17em}cm}/\text{s}\end{array}$

Therefore, wave speed$\mathrm{v}$ is $6.0\times {10}^2\text{\hspace{0.17em}cm}/\text{s}$.

As the wave travels in the negative direction of x axis, the sign is positive (+).