A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t=0 it is at position x=5 cm going towards positive x-direction. Write the equation for the displacement x at time t. Find the magnitude of the acceleration of the particle at t=4 s.
Answers
Given conditions ⇒
A = 10 cm.
T = 6 seconds.
Now, Using the Equation of the Simple Harmonic Motion,
x = ASin(ωt + Φ)
5 = 10Sin(0 + Φ)
SinΦ = 1/2
SinΦ = Sin(π/6)
When Sinθ = Sinα, then
θ = πn + (-1)ⁿα
∴ θ = α, and θ = π - α, or θ = 2π + α
∴ θ = π/6, 5π/6 or 13π/6
Now, Particle is moving in a positive x-direction. This means velocity is greater than equal to the zero.
∴ v ≥ AωCos(ωt + Φ),
ω = 2π/6 = π/3
t = 0 and A = 10 cm.
∴ 10 × π/3 Cos (0 + Φ) ≥ 0
CosΦ ≥ 0
This means Φ = π/6.
Now,
Equation of S.H.M. is,
x = ASin(ωt + Φ)
x = 10Sin(π/3 t + π/6)
____________________________
Differentiating this equation with respect to time,
v = 10π/3Cos(π/3 t + π/6)
Again differentiating it,
a = -10π²/9 × Sin(π/3 t + π/6)
At t = 4 seconds,
a = - 10.97 × Sin(4π/3+ π/6)
a = 10.97 cm/s². [since, Sin(4π/3+ π/6) = -1]
Hope it helps .
Given, r = 10cm.
At t = 0,
x = 5 cm.
T = 6 sec.
So, w = 2π/T
= 2π/6 = π/3 Sec-1
At, t = 0, x = 5 cm.
So, 5 = 10 sin (w × 0 + ∅)
= 10 sin ∅ [y = r sin wt] Sin ∅ = 1/2
⇒ ∅ = π/6
∴ Equation of displacement
x = (10cm) sin (π/3)
(ii) At t = 4 second
X = 10 sin [π/3 × 4 × π/6]
= 10 [8π+π/6] = 10 sin (3π/2)
= 10 sin (π = π/2)
= - 10 sin (π/2)
=- 10 Acceleratio
a = – w2x = – (π2/9) × (- 10) = 10.9