### Oscillations Question 9

#### Question 9 - 31 January - Shift 1

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

(1) $37.5 J$

(2) $9.75 J$

(3) $18.75 J$

(4) $12.5 J$

## Show Answer

#### Answer: (3)

#### Solution:

#### Formula: Kinetic energy of Simple harmonic motion

$u _{\max }=\frac{1}{2} m \omega^{2} A^{2}=25 J$

$KE$ at $\frac{A}{2}=\frac{1}{2} m v_1^{2}=\frac{1}{2} m \omega^{2}(A^{2}-\frac{A^{2}}{4})$

$KE=\frac{1}{2} m \omega^{2} \frac{3 A^{2}}{4}=\frac{3}{4}(\frac{1}{2} m \omega^{2} A^{2})$

$KE=\frac{3}{4} \times 25=18.75 J$