The displacement of the particle executing linear S.H.M. is $\mathrm{x}=0.25 \mathrm{sin}(11 \mathrm{t}+0.5)…
The displacement of the particle executing linear S.H.M. is $\mathrm{x}=0.25 \mathrm{sin}(11 \mathrm{t}+0.5) \mathrm{m}$. The period of S.H.M. is $\left(\pi=\frac{22}{7}\right)$
$\frac{2}{7} \mathbf{S}$
$\frac{4}{7} \mathrm{~S}$
$\frac{3}{7} \mathrm{~S}$
$\frac{1}{7}$ S
Solution
Comparing with standard equation of S.H.M.
$x=A \sin (\omega t+\phi)$
we get $\omega=11$
$\therefore T=\frac{2 \pi}{\omega}=\frac{2 \times 22}{11 \times 7}=\frac{4}{7} \kappa$
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