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प्रश्न
For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
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उत्तर १
All the waves have different phases.
The given transverse harmonic wave is:
`y (x,t)= 3.0 sin (36t + 0.018x + pi/4)` ...(i)
For x = 0, the equation reduces to:
`y (0,t) = 3.0 sin (36t + pi/4)`
Also, `omega = (2pi)/T = 36 " rad/s"^(-1)`
`:. T = pi/8 s`
Now, plotting y vs. t graphs using the different values of t, as listed in the given table.
| t(s) | 0 | T/8 | 2T/8 | 3T/8 | 4T/8 | 5T/8 | 6T/8 | 7T/8 |
| y(cm) | `(3sqrt2)/2` | 3 | `(3sqrt2)/2` | 0 | `(-3sqrt2)/2` | -3 | `(-3sqrt2)/2` | 0 |
For x = 0, x = 2, and x = 4, the phases of the three waves will get changed. This is because amplitude and frequency are invariant for any change in x. The y-t plots of the three waves are shown in the given figure.
उत्तर २
The transverse harmonic wave is
`y(x,t) = 3.0 sin (36t + 0.018x + pi/4)`
for x = 0
`y(0,t) = 3 sin(36t + 0 + pi/4) = 3 sin (36t + pi/4)` ...1
Here `omega = (2pi)/T = 36 => T =(2pi)/36`
To plot a(y) versus (t) graph, different values of y corresponding to the values of t may be tabulated as under (by making use of equation 1)
| t(s) | 0 | T/8 | 2T/8 | 3T/8 | 4T/8 | 5T/8 | 6T/8 | 7T/8 | T |
| y(cm) | `(3sqrt2)/2` | 3 | `(3sqrt2)/2` | 0 | `(-3sqrt2)/2` | -3 | `(-3sqrt2)/2` | 0 | `3/sqrt2` |
Using the values of t and y (as in the table), a graph is plotted as under The graph obtained is sinusoidal.
Similar graphs are obtained for y x = 2 cm and x = 4 cm. The (incm) oscillatory motion in the travelling wave only differs in respect of phase. Amplitude and frequency of oscillatory motion remains the same in all the cases.
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Given reasons for your answers.
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