If the amplitude of a body executing $SHM$ is doubled,what will be its energy?
(C) The energy of a body executing $SHM$ is given by the formula $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
Since $E \propto A^2$,if the amplitude $A$ is doubled $(A' = 2A)$,the new energy $E'$ becomes:
$E' = \frac{1}{2} k (2A)^2 = 4 \times (\frac{1}{2} k A^2) = 4E$.
Therefore,the energy becomes four times the original energy.
Since $E \propto A^2$,if the amplitude $A$ is doubled $(A' = 2A)$,the new energy $E'$ becomes:
$E' = \frac{1}{2} k (2A)^2 = 4 \times (\frac{1}{2} k A^2) = 4E$.
Therefore,the energy becomes four times the original energy.
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