A particle is executing simple harmonic motion with amplitude
A. When the ratio of its kinetic energy to the potential energy is , its displacement from its mean position is
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As we know,
kinetic energy
Potential energy
.
A mass of suspended from a spring of force constant executes simple harmonic oscillations. If the total energy of the oscillator is , the maximum acceleration (in ) of the mass is
5
15
45
20
Here,
In SHM,
Maximum acceleration,
A particle of mass executes simple harmonic motion with amplitude a and frequency . The average kinetic energy during its motion from the position of equilibrium to the end is
For a particle executing SHM the displacement is given by . Identify the graph which represents the variation of potential energy (P.E.) as a function of time and displacement .
I, III
II, IV
II, III
I, IV
In , the particle starts oscillating from extreme position. So at , its potential energy is maximum.
The total energy of the particle executing simple harmonic motion of amplitude A is . At a distance of from the mean position, its kinetic energy is
In , Total energy,
and, Kinetic energy,
where is the distance from the mean position.
At
As per question,
When the displacement of a particle executing simple harmonic motion is half of its amplitude, the ratio of its kinetic energy to potential energy is
Given,
from
And,
A particle undergoing simple harmonic motion has time dependent displacement given by . The ratio of kinetic to potential energy of this particle at will be :
1
2
Kinetic energy,
Potential energy,
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