Question

The amplitude of a particle executing SHM is made three-fourth keeping its time period constant. Its total energy will be

• A. $\frac{E}{2}$
• B. $\frac{3}{4}E$
• C. $\frac{\mathbf{9}}{\mathbf{16}}\mathbf{E}$
• D. None of these

Answer 1

Answer: (C)

$\frac{\mathbf{9}}{\mathbf{16}}\mathbf{E}$

Answer 2

$E = \frac{1}{2}m\omega^{2}a^{2}$⇒$\frac{E^{'}}{E} = \frac{a^{'2}}{a^{2}}$⇒$\frac{E^{'}}{E} = \frac{\left( \frac{3}{4}a \right)^{2}}{a^{2}}$ $\left( \because a^{'} = \frac{3}{4}a \right)$

⇒$E^{'} = \frac{9}{16}E$