Question

For a particle performing linear SHM, its position (x) as a function of time (t) is given by x = Asin(ωt + δ). Given that, at t = 0, particle is at +A/2 and is moving towards x = +A. Find δ

• A. π/3 rad
• B. π/6 rad
• C. π/4 rad
• D. 5π/6 rad

Hint:

The phase constant δ in SHM depends on the initial conditions (position and ve locity at t=0). Carefully consider the sign of the sine and cosine of δ to find the correct value

Answer 1

The Correct Option is B

To solve for the phase constant \delta in the equation for the particle’s position in simple harmonic motion (SHM), we start with the given equation:

x = A\sin(\omega t + \delta)

We know that when t = 0, the position x = +A/2, and the particle is moving towards +A. Let's substitute the known values into the equation:

+A/2 = A\sin(\delta)

This simplifies to:

\sin(\delta) = \frac{1}{2}

The general solution for when the sine of an angle equals \frac{1}{2} is:

\delta = \frac{\pi}{6}, \frac{5\pi}{6}, \ldots

Since the particle is moving towards +A, it indicates that it is in the first half of the cycle (0 to \pi), implying the angle is in the first quadrant. Hence, the relevant angle is:

\delta = \frac{\pi}{6}

Other options such as \frac{5\pi}{6} are incorrect in this context as they imply the particle is moving towards equilibrium from its maximum positive amplitude, which is not the initial condition given.

Therefore, the correct phase constant is:

\delta = \frac{\pi}{6} \text{ rad}