JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 23)
The velocity of a particle executing SHM varies with displacement $(x)$ as $4 v^{2}=50-x^{2}$. The time period of oscillations is $\frac{x}{7} s$. The value of $x$ is ___________. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
Answer
88
Explanation
$$4{v^2} = 50 - {x^2}$$
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 $$x = 88$$
Comments (0)
