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प्रश्न
Consider two simple harmonic motion along the x and y-axis having the same frequencies but different amplitudes as x = A sin (ωt + φ) (along x-axis) and y = B sin ωt (along y-axis). Then show that
`"x"^2/"A"^2 + "y"^2/"B"^2 - (2"xy")/"AB" cos φ = sin^2 φ`
and also discuss the special cases when
- φ = 0
- φ = π
- φ = `π/2`
- φ = `π/2` and A = B
- φ = `π/4`
Note: when a particle is subjected to two simple harmonic motions at right angle to each other the particle may move along different paths. Such paths are called Lissajous figures.
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उत्तर
Given: x = A sin (ωt + φ) ..................(1)
y = B sin ωt .............(2)
In equation (1) use,
sin (A – B) – sin A cos B + cos A sin B
x – A sin ωt. cos (φ) + A cos ωt. sin φ
x – A sin cot. cos φ = A cos cot sin φ
squaring on both sides we get,
(x – A sin cot. cos φ)2 = A2 cos2 cot sin2 φ ........(3)
In equation (3) sin at can be re-written as, `"y"/"B"` .............[from equation (2)]
Also, use
cos2 cot = 1 – sin2 ωt in equation (3)
∴ Equation (3) becomes on expansion
`("x" - "A""y"/"B". cos Φ)^2 = "A"^2 (1 - "y"^2/"B"^2) sin^2 Φ`
`"x"^2 + ("A"^2"y"^2)/"B"^2 cos^2 Φ - (2"xAy")/"B" cos Φ`
= `"A"^2 sin^2 Φ - ("A"^2"y"^2)/"B"^2 sin^2 Φ` ......(4)
`"x"^2 + ("A"^2"y"^2)/"B"^2 (sin^2 Φ + cos^2 Φ) - (2"xyA")/"B" cos Φ`
= A2 sin2 Φ ......(÷ by A2)
We get,
`"x"^2/"A"^2 + "y"^2/"B"^2. 1 - (2"xy")/"AB" cos Φ = sin^2 Φ` ......(5)
Hence proved.
Special cases:
a. φ = 0 in equation (5) we get,
`"x"^2/"A"^2 + "y"^2/"B"^2 - (2"xy")/"AB".1 = 0`
or `("x"/"A" - "y"/"B")^2` = 0
or `"x"/"A" = "y"/"B"`
y = `"B"/"A"."x"`
The above equation resembles the equation of a straight line passing through origin with a positive slope.
b. φ = π in equation (5)
`"x"^2/"A"^2 + "y"^2/"B"^2 + (2"xy")/"AB" = 0`
or `("x"/"A" + "y"/"B")^2` = 0
or `"x"/"A" = -"y"/"B"`
y = `-"B"/"A"."x"`
The above equation is an equation of a straight line passing through origin with a negative slope.
c. φ = `π/2` in equation (5)
The above equation of an ellipse whose centre is origin.
d. φ = `π/2` and A = B n equation (5)
`"x"^2/"A"^2 + "y"^2/"A"^2` = 1
x2 + y2 = A2
The above equation of a circle whose centre is origin.
e. φ = `π/4, cos π/4 = 1/sqrt2 = 1/sqrt2` equation (5) we get,
`"x"^2/"A"^2 + "y"^2/"A"^2 - (sqrt(2)"xy")/"AB" = 1/2`
The above equation is an equation of tilted ellipse.
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