Question

The total energy of the body executing SHM is \[E\] . Then the kinetic energy when the displacement is half of the amplitude, is:
A. \[\dfrac{{3E}}{4}\]
B. \[\dfrac{E}{4}\]
C. \[\dfrac{{3E}}{2}\]
D. \[\dfrac{{\sqrt 3 }}{4}E\]

Answer 1

Hint: We start by writing down the given information and the appropriate formulas. Then we apply the formula to find the total energy and find the value of potential energy and kinetic energy, then we apply the condition given in the question to the value and find the value of kinetic energy.

Formulas used:
The total energy of a particle undergoing a simple harmonic motion is given by the formula,
\[E = \dfrac{1}{2}m{\omega ^2}{a^2}\]
The value of kinetic energy of a body undergoing simple harmonic motion is given by the formula,
\[U = \dfrac{1}{2}m\omega \left( {{a^2} - {y^2}} \right)\]
Where \[m\] is the mass of the body, \[\omega \] is the angular frequency of the body and \[a\] is the amplitude of the motion.

Complete step by step answer:
Let us note down the given data, the total energy of the body is given as, \[E\]. Now we proceed to find the value of kinetic energy using the formula,
\[U = \dfrac{1}{2}m\omega \left( {{a^2} - {y^2}} \right)\]
Multiplying inside the bracket and assigning the right values,
\[U = \dfrac{1}{2}m\omega \left( {{a^2} - {y^2}} \right) \\
\Rightarrow U= E - \dfrac{1}{2}m{\omega ^2}{y^2}\]

Now we apply the given condition, that is, the amplitude becomes \[\dfrac{a}{4}\]. When we substitute that on the equation, we get
\[U = E - \dfrac{1}{2}m{\omega ^2}\left( {\dfrac{{{a^2}}}{4}} \right)\]
If we take the one fourth outside we get the value of total energy itself and the value becomes,
\[U = E - \dfrac{E}{4} \\
\therefore U = \dfrac{{3E}}{4}\]

Therefore, the correct answer is option A.

Note: Kinetic energy is a property that a moving body possesses. The total energy can be defined as the sum of kinetic and potential energy. The potential energy is the energy possessed by the body when it is at rest. We always measure the change in potential energy. For reference, we take potential energy at ground level to be zero and report the potential energy of objects with reference to ground level. The kinetic energy is maximum when the displacement from mean position is zero. The potential energy is maximum when the displacement is equal to the amplitude.