MCQ
A particle of mass $0.50 \mathrm{~kg}$ executes simple harmonic motion under force $\mathrm{F}=-50\left(\mathrm{Nm}^{-1}\right) \mathrm{x}$. The time period of oscillation is $\frac{x}{35} s$. The value of $x$ is . . . . .(Given $\pi=\frac{22}{7}$ )
  • A
    $21$
  • $22$
  • C
    $23$
  • D
    $24$

Answer

Correct option: B.
$22$
b
$\mathrm{m}=0.5 \mathrm{~kg}$

$\mathrm{~F}=-50(\mathrm{x})$

$\mathrm{ma}=(-50 \mathrm{x})$

$0.5 \mathrm{a}=-50 \mathrm{x}$

$\mathrm{a}=(-100 \mathrm{x})$

$\mathrm{W}^2=100 \Rightarrow(\mathrm{w}=10)$

$\mathrm{T}=\frac{2 \pi}{10}=\left(\frac{\pi}{5}\right)=\frac{22}{7 \times 15}=\left(\frac{22}{35}\right)$

$\frac{\pi}{35}=\frac{22}{35} \Rightarrow \mathrm{x}=22$

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