Question 48: Oscillations

Question

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

Accepted Answer

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$.
If we substitute these initial conditions into the equation for the displacement of the particle:
$x = A \sin(wt + \delta)$
We have:
$\frac{A}{2} = A \sin(\delta)$
Dividing both sides by $A$ gives us:
$\frac{1}{2} = \sin(\delta)$
The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).
Therefore, the correct answer is $\delta = \frac{\pi}{6}$.

Answer

$\frac{\pi}{2}$

Answer

$\frac{\pi}{3}$

Answer

$\frac{\pi}{4}$

Answer

$\frac{\pi}{6}$

Oscillations

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

Choose the correct answer: