First slide

Simple hormonic motion

Question

A point performs simple harmonic oscillation of period T and the equation of motion is given by $x = asin (ωt + \frac{π}{6})$ . After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?

Moderate

A

$\frac{T}{8}$
B

$\frac{T}{6}$
C

$\frac{T}{3}$
D

$\frac{T}{12}$

Solution

Velocity is the time derivative of displacement.
Writing the given equation of a point performing SHM
$x = a \sin (ωt + \frac{π}{6}) - - - - - - - (i)$
Differentiating Eq.(i) w.r.t time, we obtain
$v = \frac{dx}{dt} = a ω \cos (ωt + \frac{π}{6})$
It is given that v = $\frac{aω}{2}$ , so that
$\frac{aω}{2} = a ω \cos (ωt + \frac{π}{6})$
or $\frac{1}{2} = \cos (ωt + \frac{π}{6})$
or $\cos \frac{π}{3} = \cos (ωt + \frac{π}{6})$
or $ωt + \frac{π}{6} = \frac{π}{3} \Rightarrow ωt = \frac{π}{6}$
or $t = \frac{π}{6 ω} = \frac{π \times π}{6 \times 2 π} = \frac{T}{12}$
Thus, at $\frac{T}{12}$ velocity of the point will be equal to half of its maximum velocity.

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