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Question: During SHM, a particle has displacement \( x \) from mean position. If acceleration, kinetic energy ...

During SHM, a particle has displacement xx from mean position. If acceleration, kinetic energy and excess potential energy are represented by aa , kk and UU respectively, then choose the appropriate graph
(A)

(B)

(C)

(D)

Explanation

Solution

Hint
We write down the equations of kinetic energy and potential energy in a simple harmonic motion. Acceleration is then substituted into the formulas of kinetic and potential energy. These equations are compared to the graph to find which one of them is correct.
Formula used: kinetic energy k=12ω2(A2x2)k = \dfrac{1}{2}{\omega ^2}({A^2} - {x^2})
Potential energy U=12mω2x2U = \dfrac{1}{2}m{\omega ^2}{x^2}
Acceleration of particle a=ω2xa = {\omega ^2}x
Here,Mass of particle is represented by mm
Amplitude of particle is represented by AA
Displacement is represented by xx
Angular frequency is represented by ω\omega
Kinetic energy is represented by kk
Potential energy is represented by UU .

Complete step by step answer
kinetic energy of a particle in a simple harmonic motion is given by
k=12ω2(A2x2)\Rightarrow k = \dfrac{1}{2}{\omega ^2}({A^2} - {x^2})
k=12(ω2A2ax)\Rightarrow k = \dfrac{1}{2}({\omega ^2}{A^2} - ax)
Potential energy is given by
U=12mω2x2\Rightarrow U = \dfrac{1}{2}m{\omega ^2}{x^2}
U=12m×a×x\Rightarrow U = \dfrac{1}{2}m \times a \times x
Acceleration is given by a=ω2xa = {\omega ^2}x
Taking graph (A) we can see that acceleration is zero at xx equal to zero but from kinetic energy formula k=12(ω2A2ax)k = \dfrac{1}{2}({\omega ^2}{A^2} - ax) . kinetic energy is not equal to zero at acceleration zero hence (A) is incorrect
The graph of kinetic energy vs displacement should be parabolic. which is not the case in option (B), so it is incorrect.
Potential energy is zero when acceleration is zero and maximum when acceleration is maximum which is satisfied by the equation of potential energy U=12m×a×xU = \dfrac{1}{2}m \times a \times x hence option (c) is correct.
The sum of kinetic and potential energy is always equal to
\begin{array}{*{20}{l}} {\;U + K = \dfrac{1}{2}m{\omega ^2}{A^2}} \\\ { \Rightarrow U = - K + \dfrac{1}{2}m{\omega ^2}{A^2}} \end{array}
So it is a straight line with a negative slope and an intercept on the y-axis
Hence option (D) is correct.
Option (C) and (D) are correct.

Note
Simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.