The period of S.H.M. of a particle is 12 s. The phase difference between the positions at t = 3 s and t = 4 s will be

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TestHub · Accepted Answer

The phase of a particle executing simple harmonic motion at any time $t$ is given by $ϕ = ω t + ϕ_{0}$

The angular frequency $ω$ is related to the time period $T$ by $ω = \frac{2 π}{T}$

Given $T = 12$ s, we have $ω = \frac{2 π}{12} = \frac{π}{6}$ rad/s

The phase difference $Δ ϕ$ between the positions at $t_{1} = 3$ s and $t_{2} = 4$ s is

$Δ ϕ = ω (t_{2} - t_{1})$

Substituting the values, we get

$Δ ϕ = \frac{π}{6} (4 - 3) = \frac{π}{6}$ rad

Answer: $\frac{π}{6}$

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