The amplitude of a particle executing $SHM$ is made three-fourth keeping its time period constant. Its total energy will be
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(c) $E = \frac{1}{2}m{\omega ^2}{a^2}$==>$\frac{{E'}}{E} = \frac{{{{a'}^2}}}{{{a^2}}}$
==>$\frac{{E'}}{E} = \frac{{{{\left( {\frac{3}{4}a} \right)}^2}}}{{{a^2}}}$
==>$E' = \frac{9}{{16}}E$
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