Verified Solution • 2025–26 Session moderate

QJEEM11Phe28e: JEE Main 2011 Physics - Class 11 JEEM11Phe28e Solution: Two particles are executing simple harmonic motion of the same…

Two particles are executing simple harmonic motion of the same amplitude AA and frequency ω\omega along the xx -axis. Their mean position is separated by distance {X_0}\left( {{X_0} > A} \right) . If the maximum separation between them is (X0+A),\left( {{X_0} + A} \right), the phase difference between their motion is:

Multiple Choice Options

  • A π3{\pi \over 3}
  • B π4{\pi \over 4}
  • C π6{\pi \over 6}
  • D π2{\pi \over 2}

Step-by-Step Verified Solution

We know that, equation for SHM along x-axis is given by x=Asin⁡(ωt+ϕ)x = A\sin (\omega t + \phi )

Let mean position for 1st particle is at x = 0

So, the SHM equation for 1st particle,

x1=Asin⁡(ωt+ϕ1){x_1} = A\sin (\omega t + {\phi _1})

Now, as the separation between mean positions of both the particle is x 0_0

So, mean position for 2nd particle is x=x0x=x_0

Hence, the SHM equation for 2nd particle,

x2=x0+Asin⁡(ωt+ϕ2){x_2} = {x_0} + A\sin (\omega t + {\phi _2})

Now, the separation between particles would be

∣x2−x1∣=x0+Asin⁡(ωt+ϕ2)−Asin⁡(ωt+ϕ1)\left| {{x_2} - {x_1}} \right| = {x_0} + A\sin (\omega t + {\phi _2}) - A\sin (\omega t + {\phi _1})

=x0+A[sin⁡(ωt+ϕ2)−sin⁡(ωt+ϕ1)]= {x_0} + A\left[ {\sin (\omega t + {\phi _2}) - \sin (\omega t + {\phi _1})} \right]

We know,

sin⁡C−sin⁡D=2cos⁡(C+D2)sin⁡(C−D2)\sin C - \sin D = 2\cos \left( {{{C + D} \over 2}} \right)\sin \left( {{{C - D} \over 2}} \right)

Hence,

∣x2−x1∣=x0+2Acos⁡(ωt+ϕ1+ϕ22)sin⁡(ϕ2−ϕ12)\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right)\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)

For maximum separation,

Let cos⁡(ωt+ϕ1+ϕ22)=1\cos \left( {\omega t + {{{\phi _1} + {\phi _2}} \over 2}} \right) = 1

So, ∣x2−x1∣=x0+2Asin⁡(ϕ2−ϕ12)\left| {{x_2} - {x_1}} \right| = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)

Since, ∣x2−x1∣=x0+A\left| {{x_2} - {x_1}} \right| = {x_0} + A (given)

So, x0+A=x0+2Asin⁡(ϕ2−ϕ12){x_0} + A = {x_0} + 2A\sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right)

⇒sin⁡(ϕ2−ϕ12)=12\Rightarrow \sin \left( {{{{\phi _2} - {\phi _1}} \over 2}} \right) = {1 \over 2}

⇒ϕ2−ϕ12=π6\Rightarrow {{{\phi _2} - {\phi _1}} \over 2} = {\pi \over 6}

⇒ϕ2−ϕ1=π3\Rightarrow {\phi _2} - {\phi _1} = {\pi \over 3}

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