91Ó°ÊÓ

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{standing}\mathrm{wave},y(x,t)=0.040\left(\sin 5\mathrm{Ï€³\ae }\right)(\cos 40\mathrm{Ï€}t) \\ \mathrm{Angular}\mathrm{frequency},\mathrm{Ó\neg }=40\mathrm{Ï€} \\ \mathrm{Wavenumber},k=5\mathrm{Ï€} \end{gathered}$$

We can find the location of the nodes with different values of x from the condition that the amplitude of the wave must be zero at that point. We can get the angular speed and wave number and amplitude of the given wave by comparing the given equation of the wave with the general wave equation. Then, from this, we can find the period, speed of the standing wave, and amplitude of the interfered wave using the corresponding relations. From the given displacement of the wave, we can find the expression for its speed. Then, inserting the given conditions we can find the answers for parts g), h) and i).

Formula:

The angular frequency of the wave, $\mathrm{Ó\neg }=2\mathrm{Ï€´Ú}..............\left(1\right)$

The period of an oscillation,$T=\frac{1}{f}............\left(2\right)$

The velocity of a wave,$V=\frac{\mathrm{Ó\neg }}{k}............\left(3\right)$

The general expression of the wave, $y=y_m\sin (kx-\mathrm{Ó\neg }t).....\left(4\right)$

The standing wave equation is given as:

$y(x,t)=0.040\left(\sin 5\mathrm{Ï€³\ae }\right)\left(\cos 40\mathrm{Ï€}t\right)$

Now the nodes will be located where

$\sin 5\mathrm{Ï€³\ae }=0$

For the value of thefunction to be zero, the value for$5\mathrm{Ï€³\ae }$must be in the range of$0,\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€}........$

So, from this we can say that,

Location of the node with the smallest value is given at x=0:

$$\begin{gathered} 5\mathrm{Ï€³\ae }=0 \\ \mathrm{x}=\frac{0}{5\mathrm{Ï€}} \\ =0 \end{gathered}$$

Hence, the node is located at 0 m

From, the give range, the location of the node with the second smallest value for x is given as:

$$\begin{gathered} 5\mathrm{Ï€³\ae }=\mathrm{Ï€} \\ 5\mathrm{x}=1 \\ \mathrm{x}=\frac{1}{5} \\ =0.20\mathrm{m} \end{gathered}$$

Hence, the node is located at 0.20 m

From, the give range, the location of the node with the third smallest value for x is given as:

$$\begin{gathered} 5\mathrm{Ï€³\ae }=2\mathrm{Ï€} \\ 5\mathrm{x}=2 \\ \mathrm{x}=\frac{2}{5} \\ =0.40\mathrm{m} \end{gathered}$$

Hence, the location of node is 0.40 m

From equation (1), we get the ration of angular frequency and frequency as:

$\frac{\mathrm{Ó\neg }}{f}=2\mathrm{Ï€}$

But substituting the above equation in equation (2), we get the time period as:

$$\begin{gathered} \mathrm{Ó\neg }T=2\mathrm{Ï€} \\ T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ =\frac{2\mathrm{Ï€}}{40\mathrm{Ï€}} \\ =0.05\mathrm{s} \end{gathered}$$

Hence, the value of period of oscillation is 0.05 s

We have,

$y(x,t)=0.040\left(\sin 5\mathrm{Ï€³\ae }\right)\left(\cos 40\mathrm{Ï€}t\right)$

From the above superposition displacement formula of the resultant, we can write the two equations as:

$$\begin{gathered} y_1=0.020(\sin 5\mathrm{Ï€³\ae })(\cos 40\mathrm{Ï€}t) \\ {\mathrm{y}}_2=0.020\sin (5\mathrm{Ï€³\ae }+40\mathrm{Ï€}t) \end{gathered}$$

Comparing the given equation of the wave with the general equation of the standing wave we get,

From this we have,

$$\begin{gathered} k=5\mathrm{Ï€} \\ \mathrm{Ó\neg }=40\mathrm{Ï€} \end{gathered}$$

We also have from equation (iii), we get the speed of the waves is given as:

role="math" localid="1661165909881" $$\begin{gathered} V=\frac{40\mathrm{Ï€}}{5\mathrm{Ï€}} \\ =8.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed is 8.0 m/s.

Comparing the given equation of the wave with the general equation of the standing wave we get,

Amplitude of the standing wave,

$y_m=0.040\mathrm{m}$

Amplitude of the two travelling waves in terms of the Amplitude of the standing wave is given as,

role="math" localid="1661166105381" $$\begin{gathered} {\left(y_m\right)}_{s\tan dingwave}=2×{\left(y_m\right)}_{travellingwave} \\ {\left(y_m\right)}_{travellingwave}=\frac{\left({\left(y_m\right)}_{s\tan dingwave}\right)}{2} \\ =\frac{0.040}{2} \\ =0.020\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.020 m

The transverse velocity of the wave is given as:

$$\begin{gathered} u=\frac{dy}{dt} \\ =\frac{d(0.040(\sin 5\mathrm{Ï€³\ae }\left)\right(\cos 40\mathrm{Ï€³Ù}\left)\right)}{dt} \\ =-(0.040)×\left(40\mathrm{Ï€}\right)×\sin \left(5\mathrm{Ï€³\ae }\right)×\sin \left(40\mathrm{Ï€}t\right) \end{gathered}$$

The above function will be zero when$\left(40\mathrm{Ï€}t\right)=0$

For the value of thefunction to be zero, the value for$5\mathrm{Ï€³\ae }$must be in the range of$0,\mathrm{Ï€},2\mathrm{Ï€},3\mathrm{Ï€}.....$

From this we can say that, for the first time, transverse velocity, u to be zero,

$$\begin{gathered} 40\mathrm{Ï€}t=0 \\ t=0\mathrm{s} \end{gathered}$$

Hence, the first time when transverse velocity is zero is 0 s

For the second time from the given values of phases, transverse velocity to be zero,

$$\begin{gathered} 40\mathrm{Ï€}t=\mathrm{Ï€} \\ 40t=1 \\ t=\frac{1}{40} \\ =0.025\mathrm{s} \end{gathered}$$

Hence, the second time at which transverse velocity is zero is 0.025 s

For the third time from the given ranges, transverse velocity to be zero,

$$\begin{gathered} 40\mathrm{Ï€}t=2\mathrm{Ï€} \\ 40t=2 \\ t=\frac{2}{40} \\ =0.05\mathrm{s} \end{gathered}$$

Hence, the value of time is 0.05 s