A particle having a mass of 10 g oscillates according to the equation x = (2.0 cm) [(100 s^ -1 ) t + π/6] . Determine the position, the velocity, and the acceleration of the particle at t = 0 .

Correct answer: C. 1.0 cm , 1.73 m/s , -100 m/s^2 The given equation of motion is x = 2.0 (100t + π/6) cm . At t = 0 , the position is: x = 2.0 (π/6) = 2.0 × (1)/(2) = 1.0 cm The velocity is the rate of change of position: v = (dx)/(dt) = (d)/(dt) [2.0 (100t + π/6)] = 200 (100t + π/6) cm/s At t = 0 , the velocity is: v = 200 (π/6) = 200 × 3 2 = 100 3 cm/s = 3 m/s 1.73 m/s The acceleration is the rate of change of velocity: a = (dv)/(dt) = (d)/(dt) [200 (100t + π/6)] = -20000 (100t + π/6) cm/s^2 At t = 0 , the acceleration is: a = -20000 (π/6) = -20000 × (1)/(2) = -10000 cm/s^2 = -100 m/s^2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

A particle having a mass of 10 g oscillates according to the...

A particle having a mass of

10 \text{ g}

oscillates according to the equation

x = (2.0 \text{ cm}) \sin[(100 \text{ s}^{-1}) t + \pi/6]

. Determine the position, the velocity, and the acceleration of the particle at

t = 0

Options

A $2.0 \text{ cm}$ , $0 \text{ m/s}$ , $-100 \text{ m/s}^2$
B $1.0 \text{ cm}$ , $-1.73 \text{ m/s}$ , $100 \text{ m/s}^2$
C $1.0 \text{ cm}$ , $1.73 \text{ m/s}$ , $-100 \text{ m/s}^2$
D $1.73 \text{ cm}$ , $1.0 \text{ m/s}$ , $-200 \text{ m/s}^2$

Correct answer

C. $1.0 \text{ cm}$ , $1.73 \text{ m/s}$ , $-100 \text{ m/s}^2$

Step-by-step solution

The given equation of motion is

x = 2.0 \sin(100t + \pi/6) \text{ cm}

.

At

t = 0

, the position is:

x = 2.0 \sin(\pi/6) = 2.0 \times \dfrac{1}{2} = 1.0 \text{ cm}

The velocity is the rate of change of position:

v = \dfrac{dx}{dt} = \dfrac{d}{dt} [2.0 \sin(100t + \pi/6)] = 200 \cos(100t + \pi/6) \text{ cm/s}

t = 0

, the velocity is:

v = 200 \cos(\pi/6) = 200 \times \dfrac{\sqrt{3}}{2} = 100\sqrt{3} \text{ cm/s} = \sqrt{3} \text{ m/s} \approx 1.73 \text{ m/s}

The acceleration is the rate of change of velocity:

a = \dfrac{dv}{dt} = \dfrac{d}{dt} [200 \cos(100t + \pi/6)] = -20000 \sin(100t + \pi/6) \text{ cm/s}^2

t = 0

, the acceleration is:

a = -20000 \sin(\pi/6) = -20000 \times \dfrac{1}{2} = -10000 \text{ cm/s}^2 = -100 \text{ m/s}^2

Practice Simple Harmonic Motion on Quantrex Academy →

More from Simple Harmonic Motion