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What is the Kinetic Energy at the Equilibrium Point?
Let m be the mass of a particle executing simple harmonic motion, A be the amplitude and T be the time period of the motion. The particle possesses kinetic energy due to its velocity all along its path of motion except at the extremities. Again restoring force acts on the particle all along its path of motion except at the position of equilibrium. So, the work that is to be done to move the particle against the restoring force remains stored in the particle as potential energy.
Kinetic energy: At a displacement x from the position of equilibrium, the velocity of the particle, v = ω\(\sqrt{A^2-x^2}\).
So, the kinetic energy of the particle of mass m at that instant,
K = \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m \omega^2\)(A2 – x2) ……. (1)
- When the particle is just at the position of equilibrium, x = 0, the kinetic energy at that instant, K = \(\frac{1}{2} m \omega^2 A^2\) —this is the maximum value of the kinetic energy
- As the particle reaches any end of its path, x = ± A, the kinetic energy at that instant, K = 0 -this is the minimum value of the kinetic energy
So, the kinetic energy of a particle executing SHM is maximum at the position of equilibrium and zero at the two ends, i.e., at the two extremities of its path of motion.
Potential energy: When the particle is at a distance x from its position of equilibrium, the restoring force acting on
it is, F = mω2x.
Again, when the particle is just at the position of equilibrium, x = 0, the restoring force F = 0. i.e., no force acts on the particle.
So, within the displacement from 0 to x, the average force acting on the particle = \(\frac{0+m \omega^2 x}{2}\) = \(\frac{1}{2} m \omega^2 x\)
Now potential energy of the particle, U = work done to move the particle through the distance x against this average force = average force x displacement acting on the particle
= \(\frac{1}{2} m \omega^2 x \cdot x\) = \(\frac{1}{2} m \omega^2 x^2\) ……… (2)
- When the particle is just at the position of equilibrium, x = 0, the potential energy at that instant, U = 0—this is the minimum value of the potential energy
- As the particle reaches any end of its path, x = ±A, the potential energy at that instant, U = \(\frac{1}{2} m \omega^2 A^2\)—this is the maximum value of the potential energy.
Calculation of potential energy with the help of calculus: When the displacement of the particle is x from the position of equilibrium, the restoring force acting on it is, mω2x. Let dx be a further infinitesimal displacement of the
particle such that the above force acting on the particle remains constant throughout this displacement.
So, work done in moving the particle through dx = mω2x ᐧ dx
Hence, total work done in moving it from 0 to x, i.e., potential energy of the particle is,
U = \(\int_0^x\)mω2x = \(\frac{1}{2} m \omega^2 x^2\)(A2 – x2) …………. (1)
Total mechanical energy: Total energy of a particle executing simple harmonic motion,
E = K + U = \(\frac{1}{2} m \omega^2\)(A2 – x2) + \(\frac{1}{2}\)mω2x2 = \(\frac{1}{2}\)mω2A2.
Since m and ω are constants, if the amplitude A remains unchanged, then E constant, i.e., total energy of the particle does not depend on its displacement. As the particle moves away from the position of equilibrium, kinetic energy gradually gets transformed into potential energy. While returning from the extreme position towards the position of equilibrium, its potential energy gradually gets converted into kinetic energy. This conversion is in accordance with the law of conservation of energy Due to any external cause (e.g., air resistance), if the amplitude of the motion decreases, the total energy will also decrease. In that case the energy of the particle is transferred to the surroundings (e.g., different air particles).
The conversion of kinetic energy to potential energy and vice versa with the change in displacement of the particle executing SHM are shown in the following table. Using this table, the energy-displacement curve is plotted in Fig.
From the table, we see that, when the displacement of the particle is ± \(\frac{A}{\sqrt{2}}\),
then kinetic energy = potential energy = \(\frac{1}{4} m \omega^2 A^2\) = \(\frac{E}{2}\).
Numerical Examples
Example 1.
When a particle executing SHM is at a distance of 0.02 m from Its mean position, then its kinetic energy is thrice its potential energy. Calculate the amplitude of motion of the particle.
Solution:
According to the question, K = 3U
i.e., \(\frac{1}{2} m \omega^2\)(A2 – x2) = 3 × \(\frac{1}{2} m \omega^2 x^2\) or, A2 – x2 = 3x2
or, A2 = 4x2 or, A = |±2x| = 0.04 [since x = 0.02 m]
∴ Amplitude = 0.04 m.
Example 2.
When a particle executing SHM is at a distance of 0.02 m from its position of equilibrium, then its kinetic energy Is twice its potential energy. Calculate the distance from the position of equilibrium where its potential energy is twice its kinetic energy.
Solution:
In the first case, when x = 0.02 m, K = 2U
Example 3.
A particle of mass 0.2 kg is executing SHM along x-axis with a frequency of \(\frac{25}{\pi}\)Hz. If its kinetic energy is 0.5 J at x = 0.04 m, then find its amplitude of vibration?
Solution:
Example 4.
Total energy of a particle executing SHM is 3 J. A maximum force of 1.5 N acts on it. Time period and epoch of the SHM are 2s and 30° respectively. Establish the equation of this SHM and also find the mass of the particle.
Solution:
Total energy = \(\frac{1}{2} m \omega^2 A^2\) = 3 …… (1)
Maximum force = mass × maximum acceleration
= m × ω2A = 1.5 ……… (2)
Dividing (1) by (2) we get, \(\frac{1}{2}\)A = 2 or, A = 4 m
Time period, T = 2 s
∴ ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{2}\) = π rad ᐧ s-1
Epoch, α = 30° = \(\frac{\pi}{6}\)
So, the equation of SHM is x = Asin(ωt + α)
or, x = 4sin(πt + \(\frac{\pi}{6}\))m
Again from equation (2),
m = \(\frac{1.5}{\omega^2 A}\) = \(\frac{1.5}{\pi^2 \cdot 4}\) = 0.038 kg.
Example 5.
The equation of motion of a particle executing SHM is x = Asin(ωt + θ). Calculate the velocity and acceleration of the particle. If m is the mass of the particle, then what Is the maximum value of its kinetic energy?
Solution:
Given equation of SRM, x = Asin(ωt + θ) ……….. (1)
As the maximum value of cos(ωt + θ) is 1,
the maximum value of velocity, vmax = ωA
∴ Maximum kinetic energy of the particle
= \(\frac{1}{2} m v_{\max }^2\) = \(\frac{1}{2} m \omega^2 A^2\)
Example 6.
Amplitude of a particle of mass 0.1 kg executing SHM is 0.1 m. At the mean position its kinetic energy is 8 × 1o-3 J. Find the time period of its vibration. [HS ‘04]
Solution:
At the mean position, the potential energy of the particle = 0
Hence, kinetic energy of the particle at the mean position = total energy = \(\frac{1}{2} m \omega^2 A^2\)
Example 7.
An object of mass 10 kg executes SHM. Its time period and amplitude are 2 s and 1o m respectively. Find its kinetic energy when it is at a distance of
(i) 2 m and
(ii) 5 m respectively from Its position of equilibrium. Justify the two different results for (i) and (ii). [HS ‘01]
Solution:
T = 2 s, i.e., ω = \(\frac{2 \pi}{T}\) = \(\frac{2 \pi}{2}\) = π rad ᐧ s-1
Kinetic energy, K = \(\frac{1}{2} m \omega^2\)(A2 – x2)
i) When x = 2 m,
K = \(\frac{1}{2}\) × 10 × (π)2 × {(10)2 – (2)2}
= 5 × (3.14)2 × 96 = 4732.6 J.
ii) When x = 5 m,
K = \(\frac{1}{2}\) × 10 × (π)2 × {(10)2 – (5)2}
= 5 × (3.14)2 × 75 = 3697.35 J.
The velocity of a particle executing SHM becomes maximum at its mean position. As the displacement increases, i.e., as it moves away from the equilibrium position, its velocity decreases and so the kinetic energy also decreases.
Example 8.
A particle of mass m executes SHM with amplitude a and frequency n. What is the average kinetic energy of the particle during its motion from the position of equilibrium to the end? [AIEEE ‘07]
Solution:
KE in the mean position = \(\frac{1}{2}\)ma2w2
= \(\frac{1}{2}\)ma2(2πn)2
= 2π2ma2n2
KE at the end = 0
∴ Average kinetic energy = \(\frac{0+2 \pi^2 m a^2 n^2}{2}\) = π2ma2n2