Question

The maximum potential energy of a block executing simple harmonic motion is $25 J$ A is amplitude of oscillation At $A / 2$, the kinetic energy of the block is

• A. $18.75 J$
• B. $37.5 J$
• C. $9.75 J$
• D. $12.5 J$

Hint:

In simple harmonic motion, the total energy is the sum of potential and kinetic energy. The kinetic energy at any point is the difference between the total energy and potential energy.

Answer 1

The Correct Option is A

The maximum potential energy \( u_{\text{max}} \) is given as:

\[ u_{\text{max}} = \frac{1}{2} m \omega^2 A^2 = 25 \, \text{J} \]

The total energy is constant and equal to the maximum potential energy. The kinetic energy at \( x = \frac{A}{2} \) is:

\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} m \omega^2 \left( A^2 - \left( \frac{A}{2} \right)^2 \right) \]

\[ KE = \frac{1}{2} m \omega^2 \frac{3A^2}{4} = \frac{3}{4} \times \frac{1}{2} m \omega^2 A^2 = \frac{3}{4} \times 25 = 18.75 \, \text{J} \]

Thus, the kinetic energy is 18.75 J.