91Ó°ÊÓ

1. Wavelength,$\mathrm{λ}=20\mathrm{cm}\mathrm{or}0.20\mathrm{m}$
2. Vertical scale is set at,$u_{\mathit{m}}=5.0\mathrm{cm}/\mathrm{s}\mathrm{or}0.05\mathrm{m}/\mathrm{s}$

The number of waves that pass a fixed place in a unit of time is referred to as frequency. Using the relation between frequency and period in the formula of wave speed, we can find it.

Using the formula of maximum wave speed, we can find the amplitude of the wave.

Formula:

The velocity of the wave, $v=f\mathrm{λ}$ …(¾±)

The frequency of a wave, $f=1/T$ …(¾±¾±)

The maximum transverse speed, $u_m=y_m\mathrm{Ó\neg }$ …(¾±¾±¾±)

Where, f is the frequency, T is the time period ,$y_m$ is amplitude of wave and$\mathrm{Ó\neg }$ is angular speed

Using equation (ii) in equation (i), we get the velocity of the wave as:

$$\begin{gathered} v=\left(\frac{1}{T}\right)\mathrm{λ} \\ =\frac{20\mathrm{cm}}{4\mathrm{s}}(âˆ\mu T=4\mathrm{s}) \\ =5\frac{\mathrm{cm}}{\mathrm{s}} \end{gathered}$$

Therefore, the wave speed is$=5\frac{\mathrm{cm}}{\mathrm{s}}$

Using$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ in equation (iii), we get the amplitude of the wave as:

$u_m=y_m\left(2\mathrm{Ï€}f\right)$

$$\begin{gathered} \left(5.0\frac{\mathrm{cm}}{\mathrm{s}}\right)=y_m\left(2\mathrm{Ï€}\frac{1}{4\mathrm{s}}\right) \\ \left(5.0\frac{\mathrm{cm}}{\mathrm{s}}\right)=y_m\left(\frac{\mathrm{Ï€}}{2\mathrm{s}}\right) \\ {y}_m=\frac{\left(5.0\frac{\mathrm{cm}}{\mathrm{s}}\right)}{\left(\frac{\mathrm{Ï€}}{2\mathrm{s}}\right)} \\ =3.2\mathrm{cm} \end{gathered}$$

Therefore, the amplitude of wave is 3.2 cm

Using equation (ii) and T = 4s, we get the frequency of the wave as:

$\begin{array}{l}f=\frac{1}{4}\\ =0.25\mathrm{Hz}\end{array}$

Therefore, the frequency of wave is 0.25 Hz

From above information, we get

Hence, the wave function of the required case is given.