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प्रश्न
A particle is acted simultaneously by mutually perpendicular simple harmonic motions x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be ______.
पर्याय
an ellipse.
a parabola.
a circle.
a straight line.
उत्तर
A particle is acted simultaneously by mutually perpendicular simple harmonic motions x = a cos ωt and y = a sin ωt. The trajectory of motion of the particle will be a circle.
Explanation:
If two S.H.M act in perpendicular directions, then their resultant motion is in the form of a straight line or a circle or a parabola etc. depending on the frequency ratio of the two S.H.Ms and initial phase difference. These figures are called Lissajous figures.
Let the equations of two mutually perpendicular S.H.M of the same frequency be x = a1 sin ωt and y = a2 sin(ωt + Φ)
Then the general equation of the Lissajous figure can be obtained as `x^2/a_1^2 + y^2/a_2^2 - (2xy)/(a_1a_2) cos phi = sin^2 phi`
For `phi = 0^circ: x^2/a_1^2 + y^2/a_2^2 - (2xy)/(a_1a_2)` = 0
⇒ `(x/a_1 - y/a_2)^2` = 0
⇒ `x/a_1 = y/a_2`
⇒ `y = a_2/a_1 x`
This is a straight line which passes through the origin and its slope is `a_2/a_1`.
Lissajous figures in other conditions `("with" ω_1/ω_2 = 1)`
| Phase diff.(`phi)` | Equation | Figure |
| `pi/4` | `x^2/a_1^2 + y^2/a_2^2 - (sqrt(2)xy)/(a_1a_2) = 1/2` | Oblique ellipse |
| `pi/2` | `x^2/a_1^2 + y^2/a_2^2 = 1` |  |
| `pi` |
`x/a_1 + y/a_2` = 0 ⇒ `y = - a_2/a_1 x` |
Straight line |
We have to find the resultant displacement by adding x and y-components. According to the variation of x and y, the trajectory will be predicted.
Given, `x = a cos ωt = a sin(ωt + pi/2)` ......(i)
And `y = a sin ωt` ......(ii)
Here `a_1 = a_2 = a` and `phi = pi/2`
The trajectory of motion of the particle will be a circle of radius a.
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