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 is $x = A\sin (wt + \delta )$, then the value of $\delta$ will be
Solution
<p>The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$. </p>
<p>If we substitute these initial conditions into the equation for the displacement of the particle:</p>
<p>$x = A \sin(wt + \delta)$</p>
<p>We have:</p>
<p>$\frac{A}{2} = A \sin(\delta)$</p>
<p>Dividing both sides by $A$ gives us:</p>
<p>$\frac{1}{2} = \sin(\delta)$</p>
<p>The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).</p>
<p>Therefore, the correct answer is $\delta = \frac{\pi}{6}$.</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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