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a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period
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?a body following SHM on straight line path has an amplitude of oscill...
Given Data:
Amplitude of oscillation (A) = 4 cm
Distance from mean position when acceleration = velocity (x) = 2 cm
Finding Time Period:
- The acceleration of an object undergoing simple harmonic motion is given by the equation: a = -ω²x, where a is the acceleration, ω is the angular frequency, and x is the distance from the mean position.
- The velocity of the object is given by the equation: v = ω√(A² - x²), where v is the velocity and A is the amplitude of oscillation.
- Given that the acceleration is equal to the velocity when x = 2 cm, we can equate the two equations and solve for ω.
- Setting a = v at x = 2 cm, we get: -ω²(2) = ω√(16 - 4)
- Simplifying the equation gives: -2ω² = 2ω√12
- Dividing both sides by 2 gives: -ω² = ω√12
- Squaring both sides gives: ω⁴ = 12ω²
- Dividing by ω² gives: ω² = 12
- Taking the square root gives: ω = √12 = 2√3
- The time period of the oscillation is given by: T = 2π/ω
- Substituting the value of ω, we get: T = 2π/(2√3) = π/√3 seconds
Therefore, the time period of the oscillations is π/√3 seconds.
Amplitude of oscillation (A) = 4 cm
Distance from mean position when acceleration = velocity (x) = 2 cm
Finding Time Period:
- The acceleration of an object undergoing simple harmonic motion is given by the equation: a = -ω²x, where a is the acceleration, ω is the angular frequency, and x is the distance from the mean position.
- The velocity of the object is given by the equation: v = ω√(A² - x²), where v is the velocity and A is the amplitude of oscillation.
- Given that the acceleration is equal to the velocity when x = 2 cm, we can equate the two equations and solve for ω.
- Setting a = v at x = 2 cm, we get: -ω²(2) = ω√(16 - 4)
- Simplifying the equation gives: -2ω² = 2ω√12
- Dividing both sides by 2 gives: -ω² = ω√12
- Squaring both sides gives: ω⁴ = 12ω²
- Dividing by ω² gives: ω² = 12
- Taking the square root gives: ω = √12 = 2√3
- The time period of the oscillation is given by: T = 2π/ω
- Substituting the value of ω, we get: T = 2π/(2√3) = π/√3 seconds
Therefore, the time period of the oscillations is π/√3 seconds.
Community Answer
?a body following SHM on straight line path has an amplitude of oscill...
2pi i think
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?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period
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?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period.
?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about ?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?a body following SHM on straight line path has an amplitude of oscillations 4cm . the magnitude of its acceleration is equal to that of its velocity when body is 2cm away from its mean position. find time period.
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