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Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.
Correct answer is '2'. Can you explain this answer?
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Given Data:
- Time period of each particle, T = 12 sec
- Difference of their amplitudes, A2 - A1 = 12 cm
- Initial separation between the particles = A2 + A1 (as they are at their positive extreme positions)
Approach:
To find the minimum time after which the separation between the particles becomes 6 cm, we need to analyze the relative motion of the two particles. Since both particles are performing SHM along the same line, we can assume that their equations of motion are of the form:
P1: x1(t) = A1 * sin(ωt + φ1)
P2: x2(t) = A2 * sin(ωt + φ2)
where,
x1(t) and x2(t) are the displacements of P1 and P2 from their mean positions at time t,
A1 and A2 are the amplitudes of P1 and P2, respectively,
ω = 2π/T is the angular frequency,
φ1 and φ2 are the phase angles of P1 and P2, respectively.
Step 1: Finding the Phase Difference
Since both particles start from their positive extreme positions, we can assume that φ1 = φ2 = 0. Therefore, the equation of motion for each particle becomes:
P1: x1(t) = A1 * sin(ωt)
P2: x2(t) = A2 * sin(ωt)
If we subtract the equations, we get:
x2(t) - x1(t) = (A2 - A1) * sin(ωt)
At t = 0, the separation between the particles is A2 - A1. Therefore, the phase difference between them can be given as:
φ = ωt = sin^(-1)((A2 - A1)/(A2 - A1))
Step 2: Finding the Minimum Time
We want to find the time when the separation between the particles becomes 6 cm, i.e., x2(t) - x1(t) = 6 cm. Substituting the values in the equation:
6 = (A2 - A1) * sin(ωt)
Since sin(ωt) is a periodic function, we need to find the minimum value of t for which sin(ωt) = 1. The minimum value of sin(ωt) is 1 when ωt = π/2.
ωt = π/2
t = π/(2ω)
Step 3: Calculating the Minimum Time
Substituting the value of ω = 2π/T, we get:
t = π/(2 * 2π/T)
t = T/4
Given that T = 12 sec, substituting the value:
t = 12/4
t = 3 sec
Therefore, the minimum time after which the separation between the particles becomes 6 cm is 3 seconds.
- Time period of each particle, T = 12 sec
- Difference of their amplitudes, A2 - A1 = 12 cm
- Initial separation between the particles = A2 + A1 (as they are at their positive extreme positions)
Approach:
To find the minimum time after which the separation between the particles becomes 6 cm, we need to analyze the relative motion of the two particles. Since both particles are performing SHM along the same line, we can assume that their equations of motion are of the form:
P1: x1(t) = A1 * sin(ωt + φ1)
P2: x2(t) = A2 * sin(ωt + φ2)
where,
x1(t) and x2(t) are the displacements of P1 and P2 from their mean positions at time t,
A1 and A2 are the amplitudes of P1 and P2, respectively,
ω = 2π/T is the angular frequency,
φ1 and φ2 are the phase angles of P1 and P2, respectively.
Step 1: Finding the Phase Difference
Since both particles start from their positive extreme positions, we can assume that φ1 = φ2 = 0. Therefore, the equation of motion for each particle becomes:
P1: x1(t) = A1 * sin(ωt)
P2: x2(t) = A2 * sin(ωt)
If we subtract the equations, we get:
x2(t) - x1(t) = (A2 - A1) * sin(ωt)
At t = 0, the separation between the particles is A2 - A1. Therefore, the phase difference between them can be given as:
φ = ωt = sin^(-1)((A2 - A1)/(A2 - A1))
Step 2: Finding the Minimum Time
We want to find the time when the separation between the particles becomes 6 cm, i.e., x2(t) - x1(t) = 6 cm. Substituting the values in the equation:
6 = (A2 - A1) * sin(ωt)
Since sin(ωt) is a periodic function, we need to find the minimum value of t for which sin(ωt) = 1. The minimum value of sin(ωt) is 1 when ωt = π/2.
ωt = π/2
t = π/(2ω)
Step 3: Calculating the Minimum Time
Substituting the value of ω = 2π/T, we get:
t = π/(2 * 2π/T)
t = T/4
Given that T = 12 sec, substituting the value:
t = 12/4
t = 3 sec
Therefore, the minimum time after which the separation between the particles becomes 6 cm is 3 seconds.
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Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. Can you explain this answer?
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Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. 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 P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. 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 P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. Can you explain this answer?.
Two particles P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. 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 P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. 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 P1 and P2 are performing SHM along the same line about the same mean position. Initially they are at their positive extreme position. If the time period of each particle is 12 sec and the difference of their amplitudes is 12 cm then find the minimum time in seconds after which the separation between the particles become 6 cm.Correct answer is '2'. Can you explain this answer?.
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