The total energy of a particle executing SHM is E . If the amplitude is doubled and the frequency is halved, the new total energy will be

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The total energy of a particle executing SHM is given by $E = \frac{1}{2} m ω^{2} A^{2}$ .

Since $ω = 2 π f$ , we can write $E = 2 π^{2} m f^{2} A^{2}$ .

This shows that $E \propto f^{2} A^{2}$ .

Given that the new amplitude $A^{'} = 2 A$ and the new frequency $f^{'} = \frac{f}{2}$ .

The new total energy $E^{'}$ is given by $E^{'} \propto (f^{'})^{2} (A^{'})^{2}$ .

Substituting the values, we get $E^{'} \propto (\frac{f}{2})^{2} (2 A)^{2} \Rightarrow E^{'} \propto \frac{f ^{2}}{4} \times 4 A^{2} \Rightarrow E^{'} \propto f^{2} A^{2}$ .

Therefore, $E^{'} = E$ .

Answer: $E$

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