Physics · 2024
JEE Main · 9 April 2024, Shift 2 · Q57
A particle of mass 0.50 kg executes simple harmonic motion under force F=-50(N m^-1) x. The time period of oscillation is x/35 s. The value of x is…
A particle of mass $\displaystyle 0.50$ kg executes simple harmonic motion under force $\displaystyle F=-50\left(N m^{-1}\right) x$. The time period of oscillation is $\displaystyle \frac{x}{35} s$. The value of $\displaystyle x$ is
$\displaystyle \_\_\_\_$.
(Given $\displaystyle \pi=\frac{22}{7}$ )
Official answer
From NTA’s final answer key for this paper.
22
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