Question

Q1.particle oscillates with simple harmonic motion so that its displacement varies according to the expression X=(5cm) cos (2t + π/6) where X is in centimeters and t is in seconds. At t=0 find:
a. The displacement of particle;
b. it's velocity;
c. It's acceleration; and
d. Find the period and amplitude of motion.​

Answer 1

Step-by-step 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:

$\begin{gathered}\implies \dfrac{d}{dt} \:[5\:Cos(2t + \dfrac{\pi}{6})] = 5 \times (-Sin(2t + \dfrac{\pi}{6}) \times 2\\\\\\\implies \boxed{v = - 10 \:Sin ( 2t + \dfrac{\pi}{6})}\end{gathered}$

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:

$\begin{gathered}\implies \dfrac{d}{dt} [-10\:Sin(2t + \dfrac{\pi}{6})] = -10 \times Cos(2t + \dfrac{\pi}{6}) \times 2\\\\\\\implies \boxed{a = -20\:Cos(2t + \dfrac{\pi}{6})}\end{gathered}$

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.

Answer 2

Answer:

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