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The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at -

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प्रश्न

The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to `1/3`​ of its amplitude from its mean position is ______.

विकल्प

  • 12 × 103 J

  • 25 × 10-3J

  • 0.48 × 10-3 J

  • 0.24 × 10-3 J

MCQ
रिक्त स्थान भरें

उत्तर

The displacement of a particle of mass 3 g executing simple harmonic motion is given by Y = 3 sin (0.2 t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to `1/3`​ of its amplitude from its mean position is 0.48 × 10-3 J.

Explanation:

Equation of SHM Y = 3 sin (0.2 t) Comparing with Y = a sin ωt,

we have

a = 3m,

ω = 0.2s−1

Mass of the particle = 3g = 3 × 10−3 kg

Therefore, kinetic energy of the particle is

K = `1/2"m"omega^2("a"^2-x^2)`

= `1/2xx3xx10^-3xx(0.2)^2(3^2-1^2)` `(∵ x = "a"/3)`

= 0.48 × 10-3 J

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Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.
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