The displacements of two particles executing S.H.M. on the same line are given as y1=a sin ((π/2) t+φ) and y2=b sin ((2 π/3) t+φ). The phase difference between them at t=1 s is

Question

The displacements of two particles executing S.H.M. on the same line are given as y1=a sin ((π/2) t+φ) and y2=b sin ((2 π/3) t+φ). The phase difference between them at t=1 s is (A) π (B) (π/2) (C) (π/4) (D) (π/6)

Tardigrade Admin · Accepted Answer

Phase difference between them is just difference in the angular values. Phase difference =((2 π/3)+φ)-((π/2)+φ)=(π/6)

Q. The displacements of two particles executing S.H.M. on the same line are given as $y_{1} = a sin (\frac{π}{2} t + ϕ)$ and $y_{2} = b sin (\frac{2 π}{3} t + ϕ)$ . The phase difference between them at $t = 1 s$ is

$π$

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{6}$

Phase difference between them is just difference in the angular values.
Phase difference $= (\frac{2 π}{3} + ϕ) - (\frac{π}{2} + ϕ) = \frac{π}{6}$

Q. The displacements of two particles executing S.H.M. on the same line are given as y1​=asin(2π​t+ϕ) and y2​=bsin(32π​t+ϕ). The phase difference between them at t=1s is

π

2π​

4π​

6π​

Phase difference between them is just difference in the angular values. Phase difference =(32π​+ϕ)−(2π​+ϕ)=6π​

Q. The displacements of two particles executing S.H.M. on the same line are given as $y_{1} = a sin (\frac{π}{2} t + ϕ)$ and $y_{2} = b sin (\frac{2 π}{3} t + ϕ)$ . The phase difference between them at $t = 1 s$ is

$π$

$\frac{π}{2}$

$\frac{π}{4}$

$\frac{π}{6}$

Phase difference between them is just difference in the angular values.
Phase difference $= (\frac{2 π}{3} + ϕ) - (\frac{π}{2} + ϕ) = \frac{π}{6}$