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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.
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 ______
The quantity which does not vary periodically for a particle performing SHM is ______.
The frequency of oscillation of a particle of mass m suspended at the end of a vertical spring having a spring constant k is directly proportional to ____________.
The kinetic energy of a particle, executing SHM 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 time period of its oscillation is ____________.
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.?
The displacement of a particle performing S.H.M. is given by x = 10 sin (`omega"t"+ alpha`) metre. If the displacement of the particle is 5 m, then the phase of S.H.M. is ____________.
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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(g = acceleration due to gravity)
If the length of an oscillating simple pendulum is made `1/3` times at a place keeping amplitude the same, then its total energy (E) will be ______
The total energy of a particle performing S.H.M. is 'NOT' proportional to ______
The potential energy of a particle executing S.H.M is 2.5 J, when its displacement is half of amplitude. The total energy of the particle 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 performs S.H.M. of period 24 s. Three second after passing through the mean position it acquires a velocity of 2 π m/s. Its path length is ______.
`(sin45^circ=cos45^circ=1/sqrt2)`
A simple harmonic oscillator has amplitude A, angular velocity ω and mass m. Then, average energy in one time period will be ______.
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.
When a longitudinal wave propagates through a medium, the particles of the medium execute simple harmonic oscillations about their mean positions. These oscillations of a particle are characterised by an invariant ______.
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 body attached to a spring oscillates in horizontal plane with frequency ‘n’. Its total energy is ‘E’ and the spring constant is ‘K’ then the velocity in the mean position is ______.
