
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. Kinetic energy
B. Potential energy
C. Sum of kinetic energy and potential energy
D. Difference between the kinetic energy and potential energy.
Answer
164.4k+ views
Hint: The invariant physical quantities are the quantities which remain constant throughout the time interval of time. As the speed and displacement in the simple harmonic oscillation changes over the time, so kinetic energy and the potential energy changes.
Formula used:
\[v = \omega \sqrt {{A^2} - {x^2}} \]
where v is the speed of the particle at any displacement x from the mean position.
\[KE = \dfrac{1}{2}m{v^2}\]
where KE is the kinetic energy of the particle of mass m and speed v.
Complete step by step solution:
When a longitudinal wave propagates through a medium, the particles of the medium execute simple harmonic oscillations about their mean positions. The equation of the displacement is given as,
\[x = A\sin \left( {\omega t} \right)\]
Here x is the displacement of the particle at any time t. A is the maximum displacement of the particle from the mean position and \[\omega \] is the angular frequency of the simple harmonic oscillation.
The speed of the particle at any displacement x from the mean position is,
\[v = \omega \sqrt {{A^2} - {x^2}} \]
If the mass of the particle is m, then kinetic energy is,
\[KE = \dfrac{1}{2}m{v^2}\]
\[\Rightarrow KE = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) \\ \]
The force constant of is K then \[K = m{\omega ^2}\]
Then the potential energy is,
\[PE = \dfrac{{K{x^2}}}{2} \\ \]
\[\Rightarrow PE = \dfrac{{m{\omega ^2}{x^2}}}{2}\]
The sum of the kinetic energy and potential energy of the particle is the total mechanical energy,
\[T = KE + PE\]
\[\Rightarrow T = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) + \dfrac{1}{2}m{\omega ^2}{x^2} \\ \]
\[\therefore T = \dfrac{1}{2}m{\omega ^2}{A^2}\]
So, the sum of the kinetic energy and the potential energy of the particle in simple harmonic motion is constant, i.e. invariant. Hence, these oscillations of a particle are characterised by an invariant which is the sum of the kinetic energy and the potential energy.
Therefore, the correct option is C.
Note: As the particle moves away from the mean position the speed decreases and displacement increases. So, the kinetic energy decreases and potential energy increases. But at any displacement the sum of the kinetic energy and the potential energy is constant.
Formula used:
\[v = \omega \sqrt {{A^2} - {x^2}} \]
where v is the speed of the particle at any displacement x from the mean position.
\[KE = \dfrac{1}{2}m{v^2}\]
where KE is the kinetic energy of the particle of mass m and speed v.
Complete step by step solution:
When a longitudinal wave propagates through a medium, the particles of the medium execute simple harmonic oscillations about their mean positions. The equation of the displacement is given as,
\[x = A\sin \left( {\omega t} \right)\]
Here x is the displacement of the particle at any time t. A is the maximum displacement of the particle from the mean position and \[\omega \] is the angular frequency of the simple harmonic oscillation.
The speed of the particle at any displacement x from the mean position is,
\[v = \omega \sqrt {{A^2} - {x^2}} \]
If the mass of the particle is m, then kinetic energy is,
\[KE = \dfrac{1}{2}m{v^2}\]
\[\Rightarrow KE = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) \\ \]
The force constant of is K then \[K = m{\omega ^2}\]
Then the potential energy is,
\[PE = \dfrac{{K{x^2}}}{2} \\ \]
\[\Rightarrow PE = \dfrac{{m{\omega ^2}{x^2}}}{2}\]
The sum of the kinetic energy and potential energy of the particle is the total mechanical energy,
\[T = KE + PE\]
\[\Rightarrow T = \dfrac{1}{2}m{\omega ^2}\left( {{A^2} - {x^2}} \right) + \dfrac{1}{2}m{\omega ^2}{x^2} \\ \]
\[\therefore T = \dfrac{1}{2}m{\omega ^2}{A^2}\]
So, the sum of the kinetic energy and the potential energy of the particle in simple harmonic motion is constant, i.e. invariant. Hence, these oscillations of a particle are characterised by an invariant which is the sum of the kinetic energy and the potential energy.
Therefore, the correct option is C.
Note: As the particle moves away from the mean position the speed decreases and displacement increases. So, the kinetic energy decreases and potential energy increases. But at any displacement the sum of the kinetic energy and the potential energy is constant.
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