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The kinetic energy of a particle executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the tme period of its oscillation is
  • a)
    2π sec
  • b)
    20π sec
  • c)
    5π sec
  • d)
    π/5 sec
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The kinetic energy of a particle executing simple harmonic motion is 1...
Given data:
- Kinetic energy at mean position, KE = 16 J
- Amplitude, A = 25 cm = 0.25 m
- Mass of the particle, m = 5.12 kg

Calculating the time period of oscillation:
Let's start by finding the maximum kinetic energy of the particle in simple harmonic motion. At the mean position, all the kinetic energy is in the form of kinetic energy.

Maximum kinetic energy, KEmax = 2 * KE = 2 * 16 = 32 J
The maximum kinetic energy occurs at the amplitude of oscillation. Using the formula for kinetic energy in simple harmonic motion:

KEmax = (1/2) * m * ω^2 * A^2
where ω is the angular frequency and A is the amplitude. Substituting the given values:

32 = (1/2) * 5.12 * ω^2 * (0.25)^2

ω^2 = 128

ω = 8 rad/s
The time period of oscillation can be calculated using the formula:

Time period, T = 2π / ω = 2π / 8 = π/4 seconds = π/4 seconds
Therefore, the correct answer is option 'D' (π/4 seconds).
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The kinetic energy of a particle executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the tme period of its oscillation isa)2π secb)20π secc)5π secd)π/5 secCorrect answer is option 'D'. Can you explain this answer?
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The kinetic energy of a particle executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the tme period of its oscillation isa)2π secb)20π secc)5π secd)π/5 secCorrect answer is option 'D'. 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 The kinetic energy of a particle executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the tme period of its oscillation isa)2π secb)20π secc)5π secd)π/5 secCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The kinetic energy of a particle executing simple harmonic motion is 16 J when it is in its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the tme period of its oscillation isa)2π secb)20π secc)5π secd)π/5 secCorrect answer is option 'D'. Can you explain this answer?.
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