A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is x=A/2 and it moves along positive x-axis. The displacement of particle in time t is x = A sin (wt + δ ), then the value of δ will be

Solution

Physics sir · 3 steps

Put t = 0 into both x and v: position gives two angles, and the direction of motion picks one.

1. Use the position. At t = 0, x = A sin δ = A/2, so sin δ = 1/2. Two angles between 0 and 2π satisfy this: δ = π/6 and δ = 5π/6.

   v = dx/dt = Aω cos(ω t + δ) ⇒ v(0) = Aω cos δ

2. Use the direction. The particle moves along the positive x-axis, so v(0) > 0 and cos δ > 0. For δ = 5π/6, cos δ = -√3/2, which means motion towards negative x. For δ = π/6, cos δ = +√3/2, which fits.
3. So. δ = π/6, option (d).

The idea to keep. To find a phase constant, the position at t = 0 alone gives two possible angles. Always check the sign of the velocity at t = 0 to choose the right one.

Answer from the NTA final answer key. Solution by PrepWiser.