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Two particles execute simple harmonic motions of the same amplitude and frequency along the same straight line. They pass one another travelling in opposite directions, whenever their displacement is half of their amplitude. The phase difference between the two is :
- a)2π/3
- b)π
- c)π/6
- d)π/3
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Two particles execute simple harmonic motions of the same amplitude an...
We know the S.H.M. can be written as y = 1 sin ωt = a sin θ
For first we have,
For second,
a/2= a sinθ2 ⇒ θ2 = 300
Phase difference =
For first we have,
For second,
a/2= a sinθ2 ⇒ θ2 = 300
Phase difference =
Most Upvoted Answer
Two particles execute simple harmonic motions of the same amplitude an...
The phase difference between two particles executing simple harmonic motion can be determined by comparing their displacements at a given time.
Let's assume that the amplitude of the motion is A and the frequency is ω.
When the displacement of a particle is half of its amplitude (A/2), the phase angle can be determined as follows:
x1 = (A/2) * sin(ωt + φ1)
x2 = (A/2) * sin(ωt + φ2)
At the moment when the two particles pass each other, their displacements are equal in magnitude but opposite in direction. Therefore, we can set up the following equation:
(A/2) * sin(ωt + φ1) = -(A/2) * sin(ωt + φ2)
Dividing both sides by (A/2), we get:
sin(ωt + φ1) = -sin(ωt + φ2)
Since the sine function is an odd function, we can rewrite the equation as:
sin(ωt + φ1) = sin(-(ωt + φ2))
Using the trigonometric identity for the sine of a negative angle, we have:
sin(ωt + φ1) = -sin(ωt) * cos(φ2) - cos(ωt) * sin(φ2)
Comparing the coefficients of the sine function on both sides, we can equate them:
1 = -cos(φ2)
This implies that:
cos(φ2) = -1
Since the cosine function is equal to -1 at an angle of 180 degrees or π radians, we can conclude that:
φ2 = π
Therefore, the phase difference between the two particles is π or 180 degrees.
Let's assume that the amplitude of the motion is A and the frequency is ω.
When the displacement of a particle is half of its amplitude (A/2), the phase angle can be determined as follows:
x1 = (A/2) * sin(ωt + φ1)
x2 = (A/2) * sin(ωt + φ2)
At the moment when the two particles pass each other, their displacements are equal in magnitude but opposite in direction. Therefore, we can set up the following equation:
(A/2) * sin(ωt + φ1) = -(A/2) * sin(ωt + φ2)
Dividing both sides by (A/2), we get:
sin(ωt + φ1) = -sin(ωt + φ2)
Since the sine function is an odd function, we can rewrite the equation as:
sin(ωt + φ1) = sin(-(ωt + φ2))
Using the trigonometric identity for the sine of a negative angle, we have:
sin(ωt + φ1) = -sin(ωt) * cos(φ2) - cos(ωt) * sin(φ2)
Comparing the coefficients of the sine function on both sides, we can equate them:
1 = -cos(φ2)
This implies that:
cos(φ2) = -1
Since the cosine function is equal to -1 at an angle of 180 degrees or π radians, we can conclude that:
φ2 = π
Therefore, the phase difference between the two particles is π or 180 degrees.
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Community Answer
Two particles execute simple harmonic motions of the same amplitude an...
We know the S.H.M. can be written as y = 1 sin ωt = a sin θ
For first we have,
For second,
a/2= a sinθ2 ⇒ θ2 = 300
Phase difference =
For first we have,
For second,
a/2= a sinθ2 ⇒ θ2 = 300
Phase difference =
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