First slide

Simple harmonic motion

Question

A particle executing SHM of amplitude 'a' has a displacement a/2 at t = T/4 and a negative velocity. The epoch of the particle is

Moderate

A

$\frac{π}{3}$
B

$\frac{2 π}{3}$
C

$π$
D

$\frac{5 π}{3}$

Solution

$Equation of SHM = y \sin (ω t + ϕ)$
$When y = \frac{a}{2}, t = \frac{T}{4} = \frac{2 π}{4 ω} = \frac{π}{2 ω}$
$v = a ω \cos (ω t + ϕ) velocity is negative$
$\begin{array}{l} \frac{π}{2} < (ω t + ϕ) < \frac{3 π}{2} \\ \frac{a}{2} = a \sin (ω t + ϕ) \Rightarrow \sin (ω t + ϕ) = \frac{1}{2} \\ (\frac{π}{2} + ϕ) = \frac{5 π}{6} \end{array}$
$Substituting in the above equation, we get ϕ = π / 3$

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