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Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is [2011M]
- a)0
- b)2π/3
- c)π
- d)π/6
Correct answer is option 'B'. Can you explain this answer?
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Two particles are oscillating along two close parallel straight lines ...
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Two particles are oscillating along two close parallel straight lines ...
Let's consider the motion of the two particles in terms of simple harmonic motion. We can represent the motion of each particle as a cosine function:
Particle 1: x1 = A cos(ωt + φ1)
Particle 2: x2 = A cos(ωt + φ2)
Where x1 and x2 are the displacements of the particles at time t, A is the amplitude, ω is the angular frequency, and φ1 and φ2 are the phase differences for the two particles.
From the given information, we know that the particles pass each other when their displacement is half of the amplitude. We can write this as:
A/2 = A cos(φ1) - A cos(φ2)
Simplifying this equation, we get:
1/2 = cos(φ1) - cos(φ2)
Using the trigonometric identity cos(A) - cos(B) = -2sin((A + B)/2)sin((A - B)/2), we can rewrite the equation as:
1/2 = -2sin((φ1 + φ2)/2)sin((φ1 - φ2)/2)
Since the two particles have the same frequency, we know that their angular frequency ω is the same. Therefore, their time dependence is the same, and we can set ωt + φ1 = ωt + φ2:
φ1 = φ2
This means that the phase differences for the two particles are equal. Let's call this common phase difference φ:
φ1 = φ2 = φ
Now, let's rewrite the equation using this common phase difference:
1/2 = -2sin(φ/2)sin(0)
Since sin(0) = 0, we have:
1/2 = 0
This equation is not possible, so there is no solution for the common phase difference φ. Therefore, the phase difference is not defined.
The answer is (a) 0.
Particle 1: x1 = A cos(ωt + φ1)
Particle 2: x2 = A cos(ωt + φ2)
Where x1 and x2 are the displacements of the particles at time t, A is the amplitude, ω is the angular frequency, and φ1 and φ2 are the phase differences for the two particles.
From the given information, we know that the particles pass each other when their displacement is half of the amplitude. We can write this as:
A/2 = A cos(φ1) - A cos(φ2)
Simplifying this equation, we get:
1/2 = cos(φ1) - cos(φ2)
Using the trigonometric identity cos(A) - cos(B) = -2sin((A + B)/2)sin((A - B)/2), we can rewrite the equation as:
1/2 = -2sin((φ1 + φ2)/2)sin((φ1 - φ2)/2)
Since the two particles have the same frequency, we know that their angular frequency ω is the same. Therefore, their time dependence is the same, and we can set ωt + φ1 = ωt + φ2:
φ1 = φ2
This means that the phase differences for the two particles are equal. Let's call this common phase difference φ:
φ1 = φ2 = φ
Now, let's rewrite the equation using this common phase difference:
1/2 = -2sin(φ/2)sin(0)
Since sin(0) = 0, we have:
1/2 = 0
This equation is not possible, so there is no solution for the common phase difference φ. Therefore, the phase difference is not defined.
The answer is (a) 0.
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Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is [2011M]a)0b)2π/3c)πd)π/6Correct answer is option 'B'. Can you explain this answer?
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