AP EAMCET Engineering 21 Apr 2019 Shift 2 Paper

$( A )$ At mean position, The velocity of the body is maximum that is $A \omega$ $( B )$ The total energy during SHM is, $T E = \; \frac{1}{2} k A^{2}$ The kinetic energy is, $K E = \; \frac{1}{2} k (A^{2} - x^{2})$ It is given that, $K E = \; \frac{3}{4} T E$ Substitute the values in the above equation. $\; \frac{1}{2} k (A^{2} - x^{2}) = \; \frac{3}{4} \times \; \frac{1}{2} k A^{2}$ Simplify the above equation. $x = \pm \; \frac{A}{2}$ $( C )$ The total energy during SHM is, $T E = \; \frac{1}{2} k A^{2}$ The potential energy is, $P E = \; \frac{1}{2} k x^{2}$ It is given that, $P E = \; \frac{3}{4} T E$ Substitute the values in the above equation. $\; \frac{1}{2} k x^{2} = \; \frac{3}{4} \times \; \frac{1}{2} k A^{2}$ Simplify the above equation. $x = \pm \; \frac{\sqrt{3}}{2} A$ $( D )$ The acceleration is maximum at extreme position. Thus, option (3) is correct.