# Composition of Two SHMs

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• Sometimes particle is acted upon by two or more linear SHMs. In such a case, the resultant motion of the body depends on the periods, paths and the relative phase angles of the different SHMs to which it is subjected.
• Consider two SHMs having same period and parallel to each other, where a1 and a2 are amplitudes of two SHMs respectively. a1 anda2 are initial phase angle of two SHMs respectively. whose displacements are given by

x1 = a1 Sin (ωt + α1)   and x2 = a2 Sin (ωt + α2)

Resultant displacement of the particle subjected to above SHMs is given by

x  = x1 + x2

∴  x  = a1 Sin (ωt + α1)  +  a2 Sin (ωt + α2)

∴   x  = a1 [Sinωt . Cosα1 + Cosωt . Sinα1] + a2 [Sinωt . Cosα2 + Cosωt . Sinα2]

∴   x  = a1 Sinωt . Cosα1 + a1 Cosωt . Sinα1 + a2 Sinωt . Cosα2 + a2 Cosωt . Sinα2

∴   x  = a1 Sinωt . Cosα1  + a2 Sinωt . Cosα2 + a1 Cosωt . Sinα+ a2 Cosωt . Sinα2

∴   x  = Sinωt .(a1  Cosα1  + a2 Cosα2) + Cosωt . (a1 Sinα+ a2  Sinα2) ………….. (1)

Let, (a1  Cosα1  + a2 Cosα2)   = R Cos δ … (2)

(a1 Sinα+ a2  Sinα2) = R Sin δ    ……(3)

From Equations (1), (2) and (3)

x  = Sin ωt (R Cos δ)   + Cos ωt (R Sin δ)

∴   x  = R (Sin ωt  Cos δ   + Cos ωt  Sin δ)

∴   x  = R Sin (ωt + δ)  ………..(4)

Equation (4) indicates that resultant motion is also a S.H.M. along the same straight line

as two parent SHMs and of the same period and initial phase δ .

Squaring equations (2) and (3) and adding them

(R Cos δ)2+    (R Sin δ)2 =   (a1  Cosα1  + a2 Cosα2)2 +   ( a1 Sinα+ a2  Sinα2 )2

∴   R2 Cos2 δ+    R2 Sin2 δ =    a1Cosα1 + a2Cosα2 +2 a1 a2 Cos α1 Cos α2

+  a12 Sinα1 + a22Sin2α2 + 2 a1 a2 Sin α1 Sin α2

∴   R2 (Cos2 δ  +  Sin2 δ) =    a1(Cosα1 + Sinα1)+ a2(Cosα2 + Sin2α2)

+2 a1 a2 (Cos α1 Cos α2  +Sin α1 Sin α2)

∴   R2 (1) =    a1(1)+ a2(1) +2 a1 a2 Cos (α1 – α2)

∴   R2 =    a1+ a2 +2 a1 a2 Cos (α1 – α2) Dividing equation (3) by (2) From Equations (6) and (7) we can find the resultant and initial phase of resultant S.H.M.

#### Special Cases:

• Case 1: When the two SHMs are in the same phase then (α1 – α2)  =  0 If the two SHMs have the same amplitude then, a1 =  a2 = a

∴  R   =  a + a   =  2a

• Case 2: When the two SHMs are in opposite phase then, (α1 – α2)   =  π If the two SHMs have the same amplitudes then,  a1 =  a2 = a

R   =  a   –  a = 0

• Case 3: When the phase difference is (α1 – α2)   =  π / 2 If the two SHMs have the same amplitude then, a1 =  a2 = a Science > You are Here

1. Parashjyoti Nath

It is very helpful.

2. Sanket Talwekar

Easy to understand! Very helpful it is!

3. Machivenyika Billyard

This is great

4. Harsh bode

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😊😊