Question

Two simple harmonic motions are represented by the equations y₁ = 0.1 sin `(100pi"t"+pi/3)` and y₁ = 0.1 cos πt.

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is ______.

Options

• `pi/3`

• `(-pi)/6`

• `pi/6`

• `(-pi)/3`

Answer 1

Two simple harmonic motions are represented by the equations y₁ = 0.1 sin `(100pi"t"+pi/3)` and y₁ = 0.1 cos πt.

The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is `underlinebb((-pi)/6)`.

Explanation:

v₁ = `("dy"_1)/"dt"` 

= 0.1 × 100π cos `(100pi"t"+pi/3)`

v₂ = `("dy"_2)/"dt"`

= -0.1 π sin πt = 0.1 π cos `(pi"t"+pi/2)`

∴ Phase diff. = Φ₁ - Φ₂

= `pi/3-pi/2`

= `(2pi-3pi)/6`  

= `-pi/6`