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Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is :
- a)T/6
- b)5T/6
- c)T/3
- d)T/4
Correct answer is option 'A'. Can you explain this answer?
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Two particles are performing simple harmonic motion in a straight line...
Given information:
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are the same and equal to A and T, respectively. At time t=0, one particle has a displacement A while the other one has a displacement -A/2, and they are moving towards each other.
Approach:
To find the time at which the particles cross each other, we need to consider the equation of motion for simple harmonic motion and use the given information about their initial displacements.
Solution:
1. Equation of motion for simple harmonic motion:
The equation of motion for simple harmonic motion is given by:
x = A * sin(ωt + φ)
Where:
- x is the displacement of the particle from the equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency (ω = 2π/T, where T is the time period),
- t is the time, and
- φ is the phase constant.
2. Initial displacements:
At time t=0, one particle has a displacement A while the other one has a displacement -A/2. This means their respective equations of motion are:
Particle 1: x₁ = A * sin(φ)
Particle 2: x₂ = -(A/2) * sin(φ)
3. Time at which the particles cross each other:
To find the time at which the particles cross each other, we need to find the value of t for which x₁ = x₂.
Setting x₁ = x₂, we have:
A * sin(φ) = -(A/2) * sin(φ)
Dividing both sides by sin(φ) (assuming sin(φ) ≠ 0), we get:
A = -A/2
Simplifying, we have:
2A = -A
Dividing both sides by A (assuming A ≠ 0), we get:
2 = -1
This equation does not have any valid solution, which means that the particles do not cross each other.
Therefore, the correct answer cannot be determined from the given information.
Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are the same and equal to A and T, respectively. At time t=0, one particle has a displacement A while the other one has a displacement -A/2, and they are moving towards each other.
Approach:
To find the time at which the particles cross each other, we need to consider the equation of motion for simple harmonic motion and use the given information about their initial displacements.
Solution:
1. Equation of motion for simple harmonic motion:
The equation of motion for simple harmonic motion is given by:
x = A * sin(ωt + φ)
Where:
- x is the displacement of the particle from the equilibrium position,
- A is the amplitude of the motion,
- ω is the angular frequency (ω = 2π/T, where T is the time period),
- t is the time, and
- φ is the phase constant.
2. Initial displacements:
At time t=0, one particle has a displacement A while the other one has a displacement -A/2. This means their respective equations of motion are:
Particle 1: x₁ = A * sin(φ)
Particle 2: x₂ = -(A/2) * sin(φ)
3. Time at which the particles cross each other:
To find the time at which the particles cross each other, we need to find the value of t for which x₁ = x₂.
Setting x₁ = x₂, we have:
A * sin(φ) = -(A/2) * sin(φ)
Dividing both sides by sin(φ) (assuming sin(φ) ≠ 0), we get:
A = -A/2
Simplifying, we have:
2A = -A
Dividing both sides by A (assuming A ≠ 0), we get:
2 = -1
This equation does not have any valid solution, which means that the particles do not cross each other.
Therefore, the correct answer cannot be determined from the given information.
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Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is :a)T/6b)5T/6c)T/3d)T/4Correct answer is option 'A'. Can you explain this answer?
Question Description
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Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is :a)T/6b)5T/6c)T/3d)T/4Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is :a)T/6b)5T/6c)T/3d)T/4Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t=0 one particle has displacement A while the other one has displacement -A/2 and they are moving towards each other. If they cross each other at time t, then t is :a)T/6b)5T/6c)T/3d)T/4Correct answer is option 'A'. Can you explain this answer?.
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