A particle executes simple harmonic oscillation with an amplitude A . The period of oscillation is T . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Question

A particle executes simple harmonic oscillation with an amplitude A . The period of oscillation is T . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is (A) T/8 (B) T/12 (C) T/2 (D) T/4

Tardigrade Admin · Accepted Answer

Using the equation of SHM, x(t)=Asin(ω t) (from the equilibruim position) At x(t)=A/2 ∴ (A/2)=Asin(ω t) sin((π /6))=sin(ω t) textor,(π /6)=(2 π t/T)[∵ ω = (2 π /T)] or, t=T/12 .

Q. A particle executes simple harmonic oscillation with an amplitude $A$ . The period of oscillation is $T$ . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

$T /8$

$T /12$

$T /2$

$T /4$

Using the equation of SHM,
$x (t) = A s in (ω t)$ (from the equilibruim position)
At $x (t) = A /2$
$∴ \frac{A}{2} = A s in (ω t)$
$s in (\frac{π}{6}) = s in (ω t) or, \frac{π}{6} = \frac{2 π t}{T} [∵ ω = \frac{2 π}{T}]$
or, $t = T /12$ .

Q. A particle executes simple harmonic oscillation with an amplitude A . The period of oscillation is T . The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

T/8

T/12

T/2

T/4