A particle is executing shm withh amplitude a and has maximum velocity $v_0$ .its speed at displacement $3a/4$ will be?

$\begin{array}{l} { V_{ \max } }={ V_{ 0 } }=A\omega \\ V=\omega \sqrt { { A^{ 2 } }-{ x^{ 2 } } } \\ =\omega \sqrt { { A^{ 2 } }-{ { \left( { \dfrac { { 3A } }{ 4 } } \right) }^{ 2 } } } \\ =\omega \sqrt { { A^{ 2 } }-\dfrac { { 9{ A^{ 2 } } } }{ { 16 } } } \\ =\omega \sqrt { \dfrac { { 16{ A^{ 2 } }-9{ A^{ 2 } } } }{ { 16 } } } \\ =\omega \sqrt { \dfrac { { 7{ A^{ 2 } } } }{ { 16 } } } \\ =\dfrac { { \omega A } }{ 4 } \sqrt { 7 } \\ =\dfrac { { { v_{ 0 } }\sqrt { 7 } } }{ 4 } \end{array}$

Hence, Option $A$ is correct .