A particle executes S.H.M. of amplitude along the -axis. At , the position of the particle is and it moves along the positive -axis. The displacement of the particle is given by then the value of will be [2023]

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Solution

Q.
A particle executes S.H.M. of amplitude $A$ along the $x$ -axis. At $t = 0$ , the position of the particle is $x = \frac{A}{2}$ and it moves along the positive $x$ -axis. The displacement of the particle is given by $x = A \sin (ω t + δ),$ then the value of $δ$ will be [2023]

1 $\frac{π}{6}$

2 $\frac{π}{3}$

3 $\frac{π}{4}$

4 $\frac{π}{2}$

Ans.
(1)

$x = A \sin (ω t + δ)$

$v = A ω \cos (ω t + δ)$

$\frac{A}{2} = A \sin (ω t + δ)$

$∴$ $v is +ve$

$At t = 0, in 1st quadrant or 4th quadrant$

$\sin δ = \frac{1}{2} \Rightarrow δ = \frac{π}{6}, \frac{5 π}{6}$ $quadrant$

$∴$ $Common solution is δ = \frac{π}{6}$

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