Question

The total energy of the particle executing simple harmonic motion of amplitude A is 100 J. At a distance of 0.707 A from the mean position, its kinetic energy is

• A. 25 J
• B. 50 J
• C. 100 J
• D. 12.5 J
• E. 70 J

Hint:

At \( x = A/\sqrt{2} \), Potential Energy equals Kinetic Energy. Each is exactly half of the Total Energy!

Answer 1

The Correct Option is B

Concept: The energy in SHM is the sum of Kinetic Energy (\( K \)) and Potential Energy (\( U \)). The Total Energy (\( E \)) is conserved throughout the motion: \[ E = K + U = \frac{1}{2}kA^2 = 100 \text{ J} \] At any displacement \( x \):
• Potential Energy: \( U = \frac{1}{2}kx^2 \)
• Kinetic Energy: \( K = E - U = \frac{1}{2}k(A^2 - x^2) \)

Step 1: Interpret the displacement value.
Given \( x = 0.707 A \). Note that \( 0.707 \approx 1/\sqrt{2} \). Therefore, \( x^2 = (A/\sqrt{2})^2 = A^2/2 \).

Step 2: Calculate Potential Energy at that point.
\[ U = \frac{1}{2}k(A^2/2) = \frac{1}{2} \left( \frac{1}{2}kA^2 \right) \] Since \( \frac{1}{2}kA^2 = 100 \text{ J} \): \[ U = \frac{1}{2} (100 \text{ J}) = 50 \text{ J} \]

Step 3: Solve for Kinetic Energy.
\[ K = E - U = 100 \text{ J} - 50 \text{ J} = 50 \text{ J} \]