functionality of operator overloading can be done without operator_ justify
Answers
Explanation:
Given Information:
Displacement Equation: x = 5 Cos ( 2t + π/6 )
To Find:
Displacement of the particle at t = 0
Velocity of the particle at t = 0
Acceleration of the particle at t = 0
Time period
Amplitude of the motion
Solution:
General Equation of an Oscillation is given as:
x = A.sin (ωt ± Ф)
where,
'A' is the Amplitude, 'ω' is the angular frequency, 't' is time and 'Ф' is the phase.
Comparing the given equation with the general equation, we get:
A = 5
ω = 2
Ф = π/6
Hence the Amplitude of the particle's motion is 5 cm.
a) To find the displacement of the particle, you can just substitute t = 0 in the given equation. Substituting t = 0, we get:
⇒ x = 5. Cos ( 2 (0) + π/6 )
⇒ x = 5 . Cos (π/6)
⇒ x = 5 × √3/2
⇒ x = 8.65 / 2
⇒ x = 4.325 cm
Hence the displacement of the particle at t = 0 is 4.235 cm.
b) Velocity of the particle can be found out by differentiating displacement with respect to time. Differentiating the equation we get:
\begin{gathered}\implies \dfrac{d}{dt} \:[5\:Cos(2t + \dfrac{\pi}{6})] = 5 \times (-Sin(2t + \dfrac{\pi}{6}) \times 2\\\\\\\implies \boxed{v = - 10 \:Sin ( 2t + \dfrac{\pi}{6})}\end{gathered}
⟹
dt
d
[5Cos(2t+
6
π
)]=5×(−Sin(2t+
6
π
)×2
⟹
v=−10Sin(2t+
6
π
)
Substituting t = 0, we get:
⇒ v = -10 × Sin ( 2 (0) + π/6 )
⇒ v = -10 × Sin (π/6)
⇒ v = -10 × 0.5 = -5 cm/s
Hence the velocity of the particle at t = 0 is -5 cm/s.
c) Acceleration of the particle can be found out by differentiating velocity with respect to time. On differentiating 'v' we get:
\begin{gathered}\implies \dfrac{d}{dt} [-10\:Sin(2t + \dfrac{\pi}{6})] = -10 \times Cos(2t + \dfrac{\pi}{6}) \times 2\\\\\\\implies \boxed{a = -20\:Cos(2t + \dfrac{\pi}{6})}\end{gathered}
⟹
dt
d
[−10Sin(2t+
6
π
)]=−10×Cos(2t+
6
π
)×2
⟹
a=−20Cos(2t+
6
π
)
Substituting t = 0, we get:
⇒ a = -20 × Cos ( 2(0) + π/6 )
⇒ a = -20 × Cos (π/6)
⇒ a = -20 × √3/2
⇒ a = -10 × 1.73
⇒ a = -17.3 cm/s²
Hence the acceleration of the particle is -17.3 cm/s².
d) According to the equation given,
→ ω = 2
→ 2πf = 2
→ πf = 1
⇒ f = 1/π
Therefore Time period of the oscillation is:
⇒ T = 1/f = 'π' seconds
Hence the Time period of the oscillation is π seconds.