91Ó°ÊÓ

1. The distance between two successive points,$\mathrm{λ}=24\text{ c³¾}$
2. At $\mathrm{t}=0\mathrm{s}$,$\mathrm{s}=0.$
3. The maximum longitudinal displacement,${\mathrm{s}}_{\mathrm{m}}=0.30\text{ c³¾}$
4. Frequency ofthesource,$\mathrm{f}=25\text{ H³ú}$
5. The displacement equation,$\mathrm{s}(\mathrm{x},\mathrm{t})={\mathrm{s}}_{\mathrm{m}}\cos (\mathrm{kx}Â\pm \mathrm{Ó\neg ³Ù})$

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,

$\mathbf{Ó\neg }\mathbf{=}\mathbf{2}\mathbf{Ï€´Ú}$ …(¾±)

The wave number of a wave,

role="math" localid="1661509157708" $\mathbf{k}\mathbf{=}\mathbf{2}\mathbf{Ï€}\mathbf{/}\mathbf{λ}$ …(¾±¾±)

The velocity of a wave,

$\mathbf{v}\mathbf{=}\mathbf{Ó\neg }\mathbf{/}\mathbf{k}$ …(¾±¾±¾±)

Given that,

$\mathrm{s}(\mathrm{x},\mathrm{t})={\mathrm{s}}_{\mathrm{m}}\cos (\mathrm{kx}Â\pm \mathrm{Ó\neg ³Ù})$

${\mathrm{s}}_{\mathrm{m}}=0.30\text{ c³¾}$

Therefore, the amplitude${\mathrm{s}}_{\mathrm{m}}$ of the wave is $0.30\text{ c³¾}$

For wave number using equation (ii), we get

$\begin{array}{c}\mathrm{k}=\frac{2\mathrm{Ï€}}{24\text{ c³¾}}\\ \mathrm{k}=0.26{\text{ c³¾}}^{\text{-1}}\end{array}$

Therefore, wave number $\mathrm{k}$is$0.26{\text{ c³¾}}^{\text{-1}}$

For angular frequency using equation (i), we get

$\begin{array}{c}\mathrm{Ó\neg }=2\mathrm{Ï€}\left(25\text{ H³ú}\right)\\ \mathrm{Ó\neg }=1.6×{10}^2\text{ r²¹»å}/\text{s}\end{array}$

Therefore, angular frequency $\mathrm{Ó\neg }$is$1.6×{10}^2\text{ r²¹»å}/\text{s}$

For wave speed using equation (iii), we get

As $\mathrm{Ó\neg }=1.6×{10}^2\text{ r²¹»å}/\text{s},\mathrm{k}=0.26{\text{ c³¾}}^{\text{-1}}$

$\begin{array}{c}\mathrm{v}=\frac{(1.6×{10}^2\text{ r²¹»å}/\text{s})}{\left(0.26{\text{ c³¾}}^{\text{-1}}\right)}\\ \mathrm{v}=6.0×{10}^2\text{ c³¾}/\text{s}\end{array}$

Therefore, wave speed$\mathrm{v}$ is $6.0×{10}^2\text{ c³¾}/\text{s}$.

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