The amplitude of a particle executing $SHM$ is made three fourth keeping its time period constant, Its total energy will be :-
A) $\dfrac{E}{2}$
B) $\dfrac{E}{2}E$
C) $\dfrac{9}{{16}}E$
D) None of these

Hint:
In simple harmonic motion the total energy of the particle depends on the amplitude of the particle doing SHM. So, we will solve this problem by using the relation between energies of the SHM and displacement of the particle.



Formula used:
The total energy of the particle doing SHM is given by
  $E = \dfrac{1}{2}m{\omega ^2}{a^2}\\$


Complete step by step solution:

The particle's kinetic energy with mass $m$, equals:
$K = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2})$
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.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know that the total energy is sum of the kinetic and potential energy, then we have:
 $E = \dfrac{1}{2}m{\omega ^2}({a^2} - {x^2}) + \dfrac{1}{2}m{\omega ^2}{x^2} \\$
  $E = \dfrac{1}{2}m{\omega ^2}{a^2}\,\,\,\,....(i) \\$
According to the question, the amplitude of a particle which is executing the simple harmonic motion by three-fourth of its time period, i.e., $a' = \dfrac{3}{4}a$
And, we have the new total energy with the three-fourth of its time period:
 $E' = \dfrac{1}{2}m{\omega ^2}{a'^2}\,\,\,...(ii)$
Compare the equation $(i)$from $(ii)$, then we have:
$\dfrac{{E'}}{E} = \dfrac{{{{a'}^2}}}{{{a^2}}}$
Substitute the value of $a'$in the obtained equation, then we obtain:
$\dfrac{{E'}}{E} = \dfrac{{{{\left( {\dfrac{3}{4}a} \right)}^2}}}{{{a^2}}} \\$
 $\Rightarrow E' = \dfrac{9}{{16}}E \\$
Thus, the correct option is: (C) $\dfrac{9}{{16}}E$




Note:
 It should be noted that to solve this problem, we must thoroughly analyze it and use the appropriate formulas. In addition to applying the proper formula to determine the particle's position, we must be aware of the circumstances under which the particle will travel at its maximum speed.