A block attached to an ideal spring oscillates with amplitude $A$ and total mechanical energy $E$. If the total energy is doubled to $2E$, what will the new amplitude of oscillation be?

• A. $\frac{A}{2}$
• B. $A$
• C. $2A$
• D. $\sqrt{2}A$

Answer: (D)

Verified step by step guidance

1. Recall that the total mechanical energy \(E\) of a block-spring system undergoing simple harmonic motion is given by the formula \(E = \frac{1}{2} k A^2\), where \(k\) is the spring constant and \(A\) is the amplitude of oscillation.
2. Since the problem states that the total energy is doubled from \(E\) to \$2E\(, set up the equation for the new energy: \)2E = \frac{1}{2} k A_{new}^2\(, where \)A_{new}$ is the new amplitude.
3. Substitute the original energy expression \(E = \frac{1}{2} k A^2\) into the doubled energy equation to get \$2 \times \frac{1}{2} k A^2 = \frac{1}{2} k A_{new}^2$.
4. Simplify the equation to isolate \(A_{new}^2\): \(k A^2 = \frac{1}{2} k A_{new}^2\), then divide both sides by \(k\) to get \(A^2 = \frac{1}{2} A_{new}^2\).
5. Solve for \(A_{new}\) by multiplying both sides by 2 and taking the square root: \(A_{new} = \sqrt{2} A\). This shows that the new amplitude is \(\sqrt{2}\) times the original amplitude.