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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 (K) = 16 J
- Amplitude of oscillation (A) = 25 cm = 0.25 m
- Mass of the particle (m) = 5.12 kg

Formula:
- The kinetic energy of a particle in simple harmonic motion is given by the equation:
K = 0.5 * m * (2 * π / T) * A^2
where T is the time period of oscillation.

Solving for T:
- Substituting the given values into the formula:
16 = 0.5 * 5.12 * (2 * π / T) * 0.25^2
16 = 0.32 * (2 * π / T)
T = 2 * π / 0.32
T = 5π sec
Therefore, the time period of oscillation is 5π seconds, which corresponds to option 'c'.
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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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