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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
At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given the path length of S.H.M. = 10 cm.
When the displacement of a simple harmonic oscillator is half of its amplitude, its P.E. is 3 J. Its total energy is ______
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A body executing SHM has amplitude of 4 cm. What is the distance at which the body has equal value of both K.E. and P.E.?
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The total energy of a simple harmonic oscillator is proportional to ______.
The ratio of kinetic energy to the potential energy of a particle executing S.H.M. at a distance equal to (1/3)rd of its amplitude is ______.
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`(cos45^circ=1/sqrt2)`
A particle performs S.H.M. Its potential energies are 'U1' and 'U2' at displacements 'x1' and 'x2' respectively. At displacement (x1 + x2), its potential energy 'U' is ______.
A particle starts oscillating simple harmonically from its equilibrium position with time period T. At time t = T/12, the ratio of its kinetic energy to potential energy is ______.
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A particle executes SHM with an amplitude of 10 cm and frequency 2 Hz. At t = 0, the particle is at a point, where potential energy and kinetic energy are same. The equation of displacement of particle is ______.
A body of mass 0.5 kg performs SHM with amplitude 3 cm and force constant 10 N/m. Find its total energy.
