Q92P - Fundamentals Of Physics

Question

Two waves,

$$\begin{gathered} {\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{50}\mathit{m}\mathit{m}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left[}\mathbf{(}\mathbf{25}\mathbf{.}\mathbf{1}\mathbf{r}\mathbf{a}\mathbf{d}\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{x}\mathbf{-}\mathbf{(}\mathbf{440}\mathbf{r}\mathbf{a}\mathbf{d}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{t}\mathbf{\right]}\mathbf{ }\mathbf{and}\mathbf{ } \\ {\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{50}\mathit{m}\mathit{m}\mathbf{)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left[}\mathbf{(}\mathbf{25}\mathbf{.}\mathbf{1}\mathbf{r}\mathbf{a}\mathbf{d}\mathbf{/}\mathbf{m}\mathbf{)}\mathbf{x}\mathbf{+}\mathbf{(}\mathbf{440}\mathbf{r}\mathbf{a}\mathbf{d}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{t}\mathbf{\right]} \end{gathered}$$

travel along a stretched string. (a) Plot the resultant wave as a function of t for, $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathit{\lambda }\mathbf{/}\mathbf{8}\mathbf{,}\mathbf{ }\mathit{\lambda }\mathbf{/}\mathbf{4}\mathbf{,}\mathbf{ }\mathbf{3}\mathit{\lambda }\mathbf{/}\mathbf{8}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }\mathit{\lambda }\mathbf{/}\mathbf{2}\mathbf{ }\mathbf{where}\mathbf{ }\mathit{\lambda }$ is the wavelength. The graphs should extend from t = 0 to a little over one period. (b) The resultant wave is the superposition of a standing wave and a traveling wave. In which direction does the traveling wave move? (c) How can you change the original waves so the resultant wave is the superposition of standing and traveling waves with the same amplitudes as before but with the traveling wave moving in the opposite direction? Next, use your graphs to find the place at which the oscillation amplitude is (d) maximum and (e) minimum. (f) How is the maximum amplitude related to the amplitudes of the original two waves? (g) How is the minimum amplitude related to the amplitudes of the original two waves?

Step-by-Step Solution

Answer

1. Plot resultant wave as a function of for $x=0,\frac{\lambda }{8},\frac{\lambda }{4},\frac{3\lambda }{8},\frac{\lambda }{2}$
2. The direction of the travelling wave is along the +x axis.
3. To change the original wave to the resultant wave the amplitudes of the waves should be interchanged.
4. The place at which the oscillation amplitude is maximum is 62.6 mm
5. The place at which the oscillation amplitude is minimum is 125 mm
6. The maximum amplitude related to the amplitudes of the original two waves is related by $y_{1m}+y_{2m}$
7. The minimum amplitude related to the amplitudes of the original two waves is related by $y_{1m}-y_{2m}$

Step 1: Identification of given data

The wave equations are given as:

$$\begin{gathered} y_1=(2.50 mm) \sin \left(\left(25.1\frac{rad}{m}\right)x\right)-\sin \left(\left(440\frac{rad}{s}\right)t\right) \\ {y}_2=(1.50 mm) \sin \left(\left(25.1\frac{rad}{m}\right)x\right)\sin \left(\left(440\frac{rad}{s}\right)t\right) \end{gathered}$$

Step 2: Significance of the wave equation

A wave is an oscillation that moves through space while also transferring energy. Without moving the particles of the medium, the waves' motion results in an energy transfer.

We can plot the resultant wave as a function of t for different x using given equations of waves. The direction of the resultant wave can be concluded from the direction of the larger amplitude wave. We can find the places at which the amplitude is maximum and minimum from the wavelength observing the graph. Using the answers from parts d) and e), we can find the relations of amplitudes of the original two waves with the maximum amplitude and minimum amplitude.

 

Formula: 

The wavelength of the wave, $\lambda =\frac{2\mathrm{\pi }}{k}$                                                                           …(i)

Where, k is the wave number

Step 3: (a) Plotting the resultant wave equation

The plot for x = 0 is as below:

The plot for $x=\frac{\lambda }{8}$ is as below:

The plot for $x=\frac{\lambda }{4}$ is as below:

The plot for $x=\frac{3\lambda }{8}$is as below:

The plot for $x=\frac{\lambda }{2}$is as below:

Step 4: (b) Calculation of direction of the traveling wave

We can consider that the wave y₁ is made of two waves, one of which is a travelling wave. From the equation for  y_1, we can conclude that the direction of the travelling wave is positive x.

Step 5: (c) Checking the behavior of amplitudes

Since y₁ has a larger amplitude, the resultant wave has the direction of y₁. So, to change the original wave to the resultant wave the amplitudes of the waves should be interchanged.

Step 6: (d) Determining the place where oscillation amplitude is maximum

From the graph, we can infer that the maximum amplitude is, $y_{max}=4.0\mathrm{mm}$

Its position is

$x=\frac{\lambda }{4}..............\left(a\right)$

From the given equations

$k=25.1\frac{\mathrm{rad}}{\mathrm{m}}$

Using equation (i), the wavelength can be given as:

$$\begin{gathered} \lambda =\frac{2(3.142)}{25.1{\displaystyle \frac{\mathrm{rad}}{\mathrm{m}}}} \\ =0.250\mathrm{m} \end{gathered}$$

Substituting this value of wavelength in equation (a), we get

$$\begin{gathered} x=\frac{0.250\mathrm{m}}{4} \\ =0.0626\mathrm{m} \\ =62.6\mathrm{mm} \end{gathered}$$

Hence, the position value where oscillation amplitude is maximum is 62.6mm

Step 7: (e) Determining the place where oscillation amplitude is minimum

From the graph, we can infer that the minimum amplitude is $y_{mim}=1.0\mathrm{mm}$

Its position is given at,

$$\begin{gathered} x=\frac{\lambda }{2} \\ =\frac{0.2504\mathrm{m}}{2} \\ =0.125\mathrm{m} \\ =125\mathrm{mm} \end{gathered}$$

Hence, the position ta which the oscillation is minimum is 125mm

Step 8: (f) Determining the maximum amplitude of oscillation

We have the two amplitudes of oscillations,

$y_{1m}=2.5\mathrm{mm} \mathrm{and} y_{2m}=1.5\mathrm{mm}$

Hence, the maximum amplitude for the two given waves is achieved when,

$$\begin{gathered} y_{1m}+y_{2m}=2.5+1.5 \\ =4 \mathrm{mm} \\ ={\mathrm{y}}_{\max } \end{gathered}$$

Hence, the maximum amplitude for the oscillation of two waves is related by the formula $y_{1m}+y_{2m}$

Step 9: (g) Determining the minimum amplitude of oscillation

We have the two amplitudes of the given waves,

$y_{1m}=2.5\mathrm{mm} \mathrm{and} y_{2m}=1.5 \mathrm{mm}$

Hence, the minimum amplitude for the two given waves is achieved when

$$\begin{gathered} y_{1m}-y_{2m}=2.5-1.5 \\ =1\mathrm{mm} \\ =y_{min} \end{gathered}$$

Hence, the minimum amplitude for the oscillation of two waves is related by the formula$y_{1m}-y_{2m}$