Question

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

• A. \(E/4\)
• B. \(E\)
• C. \(2E\)
• D. \(4E\)

Hint:

Total energy of SHM is proportional to the square of the amplitude and the square of the frequency.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The total energy of a particle in SHM is \(E=\frac{1}{2}m\omega^{2}A^{2}=2\pi^{2}mf^{2}A^{2}\).

Step 2: Changes
The amplitude becomes \(2A\) and the frequency becomes \(\frac{f}{2}\).

Step 3: Calculation
\(E'=2\pi^{2}m\left(\frac{f}{2}\right)^{2}(2A)^{2}=2\pi^{2}mf^{2}A^{2}\times\frac{4}{4}=E\).

Step 4: Compare the options
The energy is unchanged, option (B).

Final Answer:
The total energy remains E. \[ \boxed{\text{(B) }E} \]