
A body executes simple harmonic motion. The potential energy (PE), kinetic energy (KE) and total energy (TE) are measured as a function of displacement $x$. Which of the following statements is true?
A) TE is zero when $x = 0$
B) PE is maximum when $x = 0$
C) KE is maximum when $x = 0$
D) KE is maximum when $x$is maximum
Answer
161.4k+ views
Hint:
To find the values of energies of a particle doing SHM on its various positions asked in the question we can use the concept of energies of a particle doing SHM.
Formula used:
The potential energy
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
$Kinetic\,energy,\,K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Complete step by step solution:
In the question, we have given that the potential, Kinetic and total energy are measured as $x$,
In order to know that a basic harmonic motion is an oscillating motion that is under a retarding force which is proportional to the degree of the displacement from an equilibrium position and the object's path must be straight, and a restoring force will be applied to the equilibrium point or mean position.
Kinetic energy is the energy that an object has as a result of the motion and is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body keeps the kinetic energy it acquired throughout its acceleration unless its speed changes. While potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$Kinetic\,energy,\,K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Kinetic energy is maximum when $x = 0$, then we have: ${K_{\max }} = \dfrac{1}{2}m{\omega ^2}{a^2}$
At $x = 0$, kinetic energy is maximum and potential energy is minimum
Similarly, potential energy when $x = 0$, we have:
$U = \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$\Rightarrow U = 0 \\$
Therefore, kinetic energy will be maximum, potential energy will be minimum and the total energy is always constant.
Thus, the correct option is: (C) KE is maximum when $x = 0$
Note:
It should be noted that the kinetic energy and the potential energy are equivalent to each other. The total energy in the simple harmonic motion which is entirely kinetic in the middle and entirely potential in the extreme positions. And an energy in the simple harmonic motion which is an entire amount of the energy that is a particle which holds when engaging in simple harmonic motion.
To find the values of energies of a particle doing SHM on its various positions asked in the question we can use the concept of energies of a particle doing SHM.
Formula used:
The potential energy
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
$Kinetic\,energy,\,K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Complete step by step solution:
In the question, we have given that the potential, Kinetic and total energy are measured as $x$,
In order to know that a basic harmonic motion is an oscillating motion that is under a retarding force which is proportional to the degree of the displacement from an equilibrium position and the object's path must be straight, and a restoring force will be applied to the equilibrium point or mean position.
Kinetic energy is the energy that an object has as a result of the motion and is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body keeps the kinetic energy it acquired throughout its acceleration unless its speed changes. While potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$Kinetic\,energy,\,K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
Kinetic energy is maximum when $x = 0$, then we have: ${K_{\max }} = \dfrac{1}{2}m{\omega ^2}{a^2}$
At $x = 0$, kinetic energy is maximum and potential energy is minimum
Similarly, potential energy when $x = 0$, we have:
$U = \dfrac{1}{2}m{\omega ^2}{x^2} \\$
$\Rightarrow U = 0 \\$
Therefore, kinetic energy will be maximum, potential energy will be minimum and the total energy is always constant.
Thus, the correct option is: (C) KE is maximum when $x = 0$
Note:
It should be noted that the kinetic energy and the potential energy are equivalent to each other. The total energy in the simple harmonic motion which is entirely kinetic in the middle and entirely potential in the extreme positions. And an energy in the simple harmonic motion which is an entire amount of the energy that is a particle which holds when engaging in simple harmonic motion.
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