A particle is subjected to two simple harmonic motions given by
x₁ =2.0 sin(100π t) and
x₂ =2.0 sin(120π t+π/3)
where x is in centimeter and t in second. Find the displacement of the particle at (a) t = 0.0125, (b) t = 0.025.
Answers
Answer ⇒ (a). -2.41 cm and (b). 0.27 cm.
Explanation ⇒
The displacement of the particle is given by,
x = x₁+x₂
∴ x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
(a). For time t = 0.0125
x =2.0 sin(100π × 0.0125) + 2.0 sin(120π × 0.0125 + π/3)
∴ x =2.0 sin(10π/8) +2.0 sin(12π/8 + π/3)
∴ x =2.0 [sin(5π/4) + sin(3π/2+π/3)]
∴ x = 2.0[-1/√2 + sin(11π/6)]
∴ x = -√2 + 2.0 sin(2π-π/6)
∴ x = -√2 - 2.0× 0.5
∴ x = -1.41 - 1
∴ x = -2.41 cm.
______________________________
(b) At time t =0.025 s,
∴ x =2.0 sin(100π × 0.025) + 2.0 sin(120π × 0.025 + π/3)
∴ x =2.0 [sin(10π/4) + sin(12π/4+π/3)]
∴ x =2.0 × sin(5π/2) + 2.0 sin(10π/3)
x =2.0 × 1.0 + 2.0 × sin(2π × 2 -2 π/3)
∴ x = 2.0[1 - sin2π/3]
∴ x = 0.27 cm
Hope it helps.
Answer ⇒ (a). -2.41 cm and (b). 0.27 cm.
Explanation ⇒
The displacement of the particle is given by,
x = x₁+x₂
∴ x = 2.0 sin(100π t)+2.0 sin(120π t+π/3)
(a). For time t = 0.0125
x =2.0 sin(100π × 0.0125) + 2.0 sin(120π × 0.0125 + π/3)
∴ x =2.0 sin(10π/8) +2.0 sin(12π/8 + π/3)
∴ x =2.0 [sin(5π/4) + sin(3π/2+π/3)]
∴ x = 2.0[-1/√2 + sin(11π/6)]
∴ x = -√2 + 2.0 sin(2π-π/6)
∴ x = -√2 - 2.0× 0.5
∴ x = -1.41 - 1
∴ x = -2.41 cm.