A particle performs linear S.H.M. When the displacement of the particle from mean position is 3 cm and 4 cm , corresponding velocities are 8 cm / s and 6 cm / s respectively. Its periodic time is

Question

TestHub · Accepted Answer

The velocity of a particle in S.H.M. is given by $v = ω A^{2} - x^{2}$

Squaring both sides, we get $v^{2} = ω^{2} (A^{2} - x^{2})$

For the two given positions:

$v_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2})$

$v_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2})$

Subtracting the second equation from the first:

$v_{1}^{2} - v_{2}^{2} = ω^{2} (x_{2}^{2} - x_{1}^{2})$

Substituting the given values $x_{1} = 3$ cm, $v_{1} = 8$ cm/s and $x_{2} = 4$ cm, $v_{2} = 6$ cm/s:

$8^{2} - 6^{2} = ω^{2} (4^{2} - 3^{2})$

$64 - 36 = ω^{2} (16 - 9)$

$28 = 7 ω^{2}$

$ω^{2} = 4 \Rightarrow ω = 2$ rad/s

The periodic time is $T = \frac{2 π}{ω} = \frac{2 π}{2} = π$ s

Answer: $π s$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion