Advertisements
Advertisements
Question
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.
Solution 1
Given:
Mass = m = 2 kg,
Energy = E = 40 J
The maximum speed of the body while crossing the path's centre (mean position) is Vmax, and the total energy is entirely kinetic energy.
∴ `1/2"mv"_"max"^2` = E
∴ vmax = `sqrt((2"E")/"m")=sqrt((2xx40)/2)` = 6.324 m/s
Solution 2
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)`
= `2sqrt10`
= 2 × 3.162
= 6.324 m/s
Speed of the particle while crossing the mean position is 6.324 m/s.
RELATED QUESTIONS
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 ______
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 particle starting from the mean position performs linear S.H.M. Its amplitude is 'A' and total energy is 'E'. At what displacement its kinetic energy is 3E/4?
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 ______.
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 ______.
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 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 ______.
`[sin pi/3 = cos pi/6 = sqrt3/2, sin pi/6 = cos pi/3 = 1/2]`
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.
