A particle executes S.H.M. of amplitude A along $x$-axis. At $t=0$, the position of the particle is $\mathrm{x}=\frac{\mathrm{A}}{2}$ and it moves along positive x-axis. The displacement of particle in time $t$ is $\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\delta)$, then the value $\delta$ will be :

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Elasticity & SHM

A particle executes S.H.M. of amplitude A along

x

-axis. At

t = 0

, the position of the particle is

x = \frac{A}{2}

and it moves along positive x-axis. The displacement of particle in time

t

x = A sin (ω t + δ)

, then the value

δ

will be :

Answer: (A)

t = 0

x = \frac{A}{2} = A sin δ

, so

sin δ = \frac{1}{2}

and

δ = \frac{π}{6}

\frac{5 π}{6}

. The particle moves along

+ x

, so the velocity

v = A ω cos δ

must be positive, which requires

cos δ > 0

. Only

δ = \frac{π}{6}

meets both conditions.

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option