Question

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

1. φ = 0
2. φ = π
3. φ = `π/2`
4. φ = `π/2` and A = B
5. 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.

Answer 1

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.