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Discuss the composition of two S.H.M.s along the same path having same period. Find the resultant amplitude and intial phase.

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#### Solution

Analytical treatment:

i. Let the two linear S.H.M’s be given by equations,

x_{1} = A_{1} sin (ωt + α_{1}) …(1)

x_{2} = A_{2} sin (ωt + α_{2}) …(2)

where A_{1}, A_{2} are amplitudes; α_{1}, α_{2} are initial phase angles and x_{1}, x_{2} are the displacement of two S.H.M’s in time ‘t’. ω is same for both S.H.M’s.

ii. The resultant displacement of the two S.H.M’s is given by,

x = x_{1} + x_{2} ....(3)

iii. Using equations (1) and (2) , equation (3) can be written as,

x = A_{1} sin (ωt + α_{1}) + A_{2} sin (ωt + α_{2})

= A_{1} [sin ωt cos α_{1} + cos ωt sin α_{1}] + A_{2} [sin ωt cos α_{2} + cos ωt sin α_{2}]

= A_{1} sin ωt cos α_{1} + A_{1} cos ωt sin α_{1} + A_{2} sin ωt cos α_{2} + A_{2} cos ωt sin α_{2}

= [A_{1} sin ωt cos α_{1} + A_{2} sin ωt cos α_{2}] + [A_{1} cos ωt sin α_{1} + A_{2} cos ωt sinα_{2}]

∴ x = sin ωt [A1 cos α_{1} + A_{2} cos α_{2}] + cos ωt [A_{1} sin α_{1} + A_{2} sin α_{2}] …(4)

iv. Let A_{1} cos α_{1}+ A_{2} cos α_{2} = R cos δ …(5)

and A_{1} sin α_{1} + A_{2} sin α_{2} = R sin δ …(6)

v. Using equations (5) and (6), equation (4) can be written as,

x = sin ωt. R cos δ + cos ωt.R sin δ

= R [sin ωt cos δ + cos ωt sin δ]

∴ x = R sin (ωt + δ) ....(7)

Equation (7) represents linear S.H.M. of amplitude R and initial phase angle δ with same period.

Resultant amplitude (R):

Squaring and adding equations (v) and (vi) we get,

(A_{1} cos α_{1} + A_{2} cos α_{2})^{2} + (A_{1} sin α_{1} + A_{2} sin α_{2})^{2} = R^{2}cos^{2}δ + R^{2}sin^{2}δ

**∴** A_{1}^{2}cos^{2} α1+ A_{2}^{2} cos^{2} α_{2} + 2A_{1} A_{2} cosα_{1} cosα_{2} +A_{1}^{2} sin^{2} α_{1} + A_{2}^{2}sin^{2} α_{2} + 2A_{1}A_{2} sinα_{1} sinα_{2} = R^{2} (cos^{2} δ + sin^{2} δ)

**∴** A_{1}^{2} (cos^{2} α1 + sin^{2} α_{1}) + A_{2}^{2} (cos^{2} α_{2} + sin^{2} α_{2}) + 2A_{1} A_{2} (cosα_{1} cosα_{2} + sinα_{1} sinα_{2}) = R^{2}

**∴ **A_{1}^{2} + A_{2}^{2} + 2A_{1} A_{2} cos (α_{1} - α_{2}) = R^{2}

**∴** `"R"=+-sqrt(A_1^2+A_2^2+2A_1A_2cos(alpha_1-alpha_2))` ...........(8)

Equation (8) represents resultant amplitude of two S.H.M’s.

Resultant (intial) phase (δ):

Dividing equation (6) by (5), we get

`(A_1sinalpha_1 +A_2sinalpha_2)/(A_1cosalpha_1+A_2cosalpha_2)=(Rsindelta)/(Rcosdelta)`

`therefore(A_1sinalpha_1 +A_2sinalpha_2)/(A_1cosalpha_1+A_2cosalpha_2)=tandelta`

`thereforedelta=tan^(-1)[(A_1sinalpha_1 +A_2sinalpha_2)/(A_1cosalpha_1+A_2cosalpha_2)]` ..................(9)

Equation (9) represents resultant or intial phase of two S.H.M’s.

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