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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.
Deduce the expression for kinetic energy, potential energy, and total energy of a particle performing S.H.M. State the factors on which total energy depends.
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 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 ______.
A body oscillates simply harmonically with a period of 2 seconds, starting from the origin. Its kinetic energy will be 75% of the total energy after time ______
`(sin30^circ = cos60^circ = 1/2)`
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(g = acceleration due to gravity)
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 ______.
Two springs of spring constants 'K' and '2K' are stretched by same force. If 'E1' and 'E2' are the potential energies stored in them respectively, then ______.
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`(cos45^circ=1/sqrt2)`
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 ______.
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 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 ______.
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 ______.
