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MHT CET · Physics · Oscillations

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 \(\left(\pi=\frac{22}{7}\right)\)

  1. A \(\frac{2}{7} \mathbf{S}\)
  2. B \(\frac{4}{7} \mathrm{~S}\)
  3. C \(\frac{3}{7} \mathrm{~S}\)
  4. D \(\frac{1}{7}\) S
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{4}{7} \mathrm{~S}\)

Step-by-step Solution

Detailed explanation

Comparing with standard equation of S.H.M.
\(x=A \sin (\omega t+\phi)\)
we get \(\omega=11\)
\(\therefore T=\frac{2 \pi}{\omega}=\frac{2 \times 22}{11 \times 7}=\frac{4}{7} \kappa\)