QJEEM23Phbb17: JEE Main 2023 Physics - Class 11 JEEM23Phbb17 Solution: The velocity of a particle executing SHM varies with…

$4{v^2} = 50 - {x^2}$

or $v = {1 \over 2}\sqrt {50 - {x^2}}$

Comparing the above equation with $v = \omega \sqrt {{A^2} - {x^2}}$

$\Rightarrow \omega = {1 \over 2}$

& $A = \sqrt {50}$

so ${{2\pi } \over T} = {1 \over 2}$

$\Rightarrow T = 4\pi \sec$

$= 4 \times {{22} \over 7}\sec$

$T = {{88} \over 7}\sec$

so $x = 88$