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प्रश्न
The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the center of the path.
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उत्तर
Given:
m = 2 kg,
T.E. = 40 J
To find:
Speed while crossing the mean position (vmax) = ?
Formula:
T.E. = `1/2 mv_"max"^2`
Calculation:
From formula,
`v_"max" = sqrt((2 xx T.E.)/m)`
= `sqrt((2 xx 40)/2)`
= `2 sqrt 10`
= 2 × 3.162
= 6.324 m/s
The speed of the particle while crossing the mean position is 6.324 m/s.
संबंधित प्रश्न
Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for the total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
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A particle is executing Simple Harmonic Motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be ______.
A particle of mass ‘m’ is executing S.H.M. about the origin on x-axis with frequencу `sqrt ((k a)/(pi m))`, where ‘k’ is a constant and ‘a’ is the amplitude of S.H.M. If ‘x’ is a displacement of a particle, at time ‘t’, potential energy of the particle will be ______.
