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A particle undergoing SHM has velocities of $10 {cms}^{- 1} and 8 {cms}^{- 1}$ at the displacements 4 cm and 5 cm from the mean position. Its period is

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$π$

$2 π$

$\frac{3 π}{2}$

$\frac{π}{2}$

answer is C.

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Detailed Solution

$v^{2} = ω^{2} (A^{2} - x^{2})$

$10^{2} = ω^{2} (A^{2} - 4^{2})$

$8^{2} = ω^{2} (A^{2} - 5^{2})$

$10^{2} - 8^{2} = ω^{2} (25 - 16)$

$\Rightarrow ω = 2 r a d s^{- 1} ∴ T = \frac{2 π}{ω} = π$

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