91Ó°ÊÓ

• The scale ofy axis is set at ${\mathrm{y}}_{\mathrm{s}}=4.0\mathrm{cm}\mathrm{or}0.04\mathrm{m}$
• String tension, T=3.6 N
• Linear density of the string,$\mathrm{μ}=25\mathrm{g}/\mathrm{cm}\mathrm{or}0.025\mathrm{kg}/\mathrm{m}$

The maximum displacement of the particle undergoing oscillation, perpendicular to the direction of motion, is known as the amplitude of wave. The distance traveled by the wave along the axis of motion in one complete revolution, is known as wavelength of wave. The time taken by the wave to complete one oscillation is called Time Period.

Formula:

The general expression of the wave, $\mathrm{y}(\mathrm{x},\mathrm{t})={\mathrm{y}}_{\mathrm{m}}\sin \left(\mathrm{kx}Â\pm \mathrm{Ó\neg t}+\mathrm{φ}\right)$ (i)

The angular frequency of the wave, $\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{t}$ (ii)

The wave number of a wave, $k=\frac{2\mathrm{π}}{\mathrm{λ}}$ (iii)

The speed of the transverse wave, $v=\sqrt{\left(\frac{T}{\mathrm{μ}}\right)}$ (iv)

The period of a wave, $t=\frac{\mathrm{λ}}{v}$ (v)

Here, T is the tension in string, $\mathrm{λ}$is the wavelength and v is the speed of the wave.

From the graph, we can observe that the maximum displacement of the particle along y axis is the amplitude of the wave. Hence, the value of amplitude is 5 cm or 0.05 m

From the graph, by measuring the distance between two consecutive crests, we get the wavelength to be 40 cm

Using equation (iv) and the given values, we get the speed of the wave as:

$$\begin{gathered} v=\sqrt{\frac{3.6N}{0.025}} \\ =\sqrt{144} \\ =12m/s \end{gathered}$$

Hence, the wave speed of the wave is 12 m/s

Using equation (v) and the given values, we get the time period as:

$$\begin{gathered} t=\frac{0.4m}{12m/s} \\ =0.033s \end{gathered}$$

Hence, the time period of the wave is 0.033 s

The maximum transverse speed of a particle in the string can be calculated as:

$$\begin{gathered} v_y=\frac{dy}{dt} \\ =\frac{d}{dt}\left[y_m\sin \left(kxÂ\pm \mathrm{Ó\neg }t+\mathrm{φ}\right)\right] \\ =Â\pm \mathrm{Ó\neg }{y}_m\cos \left(kxÂ\pm \mathrm{Ó\neg }t+\mathrm{φ}\right) \end{gathered}$$

At maximum velocity,$\left(kxÂ\pm \mathrm{Ó\neg }t+\mathrm{φ}\right)=1$

$v_{y}max=\mathrm{Ó\neg }{y}_m$

Using equation (ii) in the above equation of maximum velocity, we get

$$\begin{gathered} v_{ymax}=\frac{2\mathrm{π}}{t}{y}_m \\ =\frac{2×3.14}{0.033s}×0.05m \\ =9.4m/s \end{gathered}$$.

Hence, the value of transverse speed of a particle is 9.4 m/s

Using equation (ii) and the given values, the wave number is given as:

$\begin{array}{l}k=\frac{2×3.14}{0.4\mathrm{m}}\\ =15.7/m\\ ≈16/m\end{array}$

Hence, the wave number is 16/m

Using equation (ii) and the given values, we get the angular frequency of the wave is given as:

$\begin{array}{l}\mathrm{Ó\neg }=\frac{2×3.14}{0.033}\\ =190/s\\ =1.9×{10}^2{s}^{-1}\end{array}$

Hence, the angular frequency is$1.9×{10}^2{s}^{-1}$

At t = 0, x = 0,the expression of the wave becomes:

$y=y_m\sin \left(\mathrm{φ}\right)$

It is mentioned that the y axis is set at$y\left(y_s\right)=0.04mandwecalculated{y}_m=0.05m$

Hence, the value of phase difference is given as:

$$\begin{gathered} \sin \mathrm{φ}=\frac{y_s}{y_m} \\ \sin \mathrm{φ}=\frac{0.04m}{0.05m} \\ \sin \mathrm{φ}=0.8 \\ \mathrm{φ}={\sin }^{-1}\left(0.8\right) \\ =0.93rad \end{gathered}$$

Phase difference is 0.93 rad.

Sign in front of the angular frequency $\mathrm{Ó\neg }$ is positive.