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प्रश्न
Obtain an expression for the resultant amplitude of, the composition of two S.H.M.’s having the same period along the same path.
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उत्तर
- Consider a particle simultaneously subjected to two S.H.M.s having the same period and along the same path (let it be along the x-axis) but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant.
- Let the two linear S.H.M’s be given by equations,
x1 = A1 sin (ωt + Φ1) ....(1)
x2 = A2 sin (ωt + Φ2) ....(2)
where A1, A2 are amplitudes; Φ1, Φ2 are initial phase angles, and x1, x2 are the displacement of two S.H.M’s in time ‘t’. ω is the same for both S.H.M’s. - The resultant displacement of the two S.H.M’s is given by,
x = x1 + x2 ....(3)
Using equations (1) and (2) , equation (3) can be written as,
x = A1 sin (ωt + Φ1) + A2 sin (ωt + Φ2)
= A1 [sin ωt cos Φ1 + cos ωt sin Φ1] + A2 [sin ωt cos Φ2 + cos ωt sin Φ2]
= A1 sin ωt cos Φ1 + A1 cos ωt sin Φ1 +A2 sin ωt cos Φ2 + A2 cos ωt sin Φ2
= [A1 sin ωt cos Φ1 + A2 sin ωt cos Φ2] + [A1 cosωt sin Φ1 + A2 cos ωt sinΦ2]
∴ x = sin ωt [A1 cos Φ1 + A2 cos Φ2] + cos ωt [A1 sin Φ1 + A2 sin Φ2] .…(4) - As A1, A2, Φ1 and Φ2 are constants, we can combine them in terms of another convenient constants R and δ as
A1 cos Φ1 + A2 cos Φ2 = R cosδ .…(5)
and A1 sin Φ1 + A2 sin Φ2 = R sin δ .…(6) - 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) is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase δ. It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M. - Resultant amplitude is,
R = `sqrt(("R" sinδ)^2 + ("R" cosδ)^2)`
Squaring and adding equations (5) and (6) we get, (A1 cos Φ1 + A2 cos Φ2)2 + (A1 sin Φ1 + A2 sin Φ2)2 = R2 cos2 δ + R2 sin2 δ
∴ `A_1^2 cos^2 Φ_1 + A_2^2 cos^2 Φ_2 + 2A_1 A_2 cos Φ_1 cosΦ_2 + A_1^2 sin^2 Φ_1 + A_2^2 sin^2 Φ_2 + 2A_1A_2 sinΦ_1 sinΦ_2 = R^2 (cos^2 δ + sin^2 δ)`
∴ `A_1^2 (cos^2 Φ_1 + sin^2 Φ_1) + A_2^2 (cos^2 Φ_2 + sin^2 Φ_2) + 2A_1A_2 (cos Φ_1 cos Φ_2 + sinΦ_1 sinΦ_2) = R^2`
∴ `A_1^2 + A_2^2 + 2A_1A_2 cos (Φ_1 − Φ_2) = R^2`
∴ R = ± `sqrt(A_1^2 + A_2^2 + 2A_1A_2cos(Φ_1 − Φ_2))` ....(8)
Equation (8) represents resultant amplitude of two S.H.M’s.
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