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Question

Two particles move parallel to x− axis about the origin with the same amplitude and frequency. At a certain instant, they are found at distance A3 from the origin on opposite sides, but their velocities are found to be in the same direction. Find the phase difference between the two particles.

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Solution

The correct option is **D** cos−1(79)

Let equations of the two SHMs be

x1=Asinωt ......(1)

x2=Asin(ωt+ϕ) .......(2)

At time t, let x1=A3,x2=−A3

⇒A3=Asinωt

& −A3=Asin(ωt+ϕ)

which gives, sinωt=13 ......(3)

& sin(ωt+ϕ)=−13 .....(4)

From Eq. (3) and (4), using sin(A+B)=sinAcosB+cosAsinB, we can write that

sinωtcosϕ+cosωtsinϕ=−13

⇒13cosϕ+√1−19sinϕ=−13

⇒√89sinϕ=−13−13cosϕ

Squaring on both sides, we get

(89)(1−cos2ϕ)=19(1+cos2ϕ+2cosϕ)

⇒cos2ϕ+2cosϕ+1+8cos2ϕ−8=0

⇒9cos2ϕ+2cosϕ−7=0

⇒9cos2ϕ+9cosϕ−7cosϕ−7=0

⇒cosϕ=(79) or −1

⇒ϕ=π or cos−1(79)

Differentiating Eqs. (1) and (2), we obtain

v1=Aωcosωt ....(5)

and v2=Aωcos(ωt+ϕ) ......(6)

If we put ϕ=π, we find v1 and v2 are of opposite signs.

Hence, ϕ=π is not acceptable.

∴ Phase difference between the two particles is ϕ=cos−1(79)

Thus, option (d) is the correct answer.

Let equations of the two SHMs be

x1=Asinωt ......(1)

x2=Asin(ωt+ϕ) .......(2)

At time t, let x1=A3,x2=−A3

⇒A3=Asinωt

& −A3=Asin(ωt+ϕ)

which gives, sinωt=13 ......(3)

& sin(ωt+ϕ)=−13 .....(4)

From Eq. (3) and (4), using sin(A+B)=sinAcosB+cosAsinB, we can write that

sinωtcosϕ+cosωtsinϕ=−13

⇒13cosϕ+√1−19sinϕ=−13

⇒√89sinϕ=−13−13cosϕ

Squaring on both sides, we get

(89)(1−cos2ϕ)=19(1+cos2ϕ+2cosϕ)

⇒cos2ϕ+2cosϕ+1+8cos2ϕ−8=0

⇒9cos2ϕ+2cosϕ−7=0

⇒9cos2ϕ+9cosϕ−7cosϕ−7=0

⇒cosϕ=(79) or −1

⇒ϕ=π or cos−1(79)

Differentiating Eqs. (1) and (2), we obtain

v1=Aωcosωt ....(5)

and v2=Aωcos(ωt+ϕ) ......(6)

If we put ϕ=π, we find v1 and v2 are of opposite signs.

Hence, ϕ=π is not acceptable.

∴ Phase difference between the two particles is ϕ=cos−1(79)

Thus, option (d) is the correct answer.

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