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 amplitude, the time period, and the spring constant for this motion.

Correct answer: B. 2.0 cm , 0.063 s , 100 N/m Comparing the given equation x = (2.0 cm) [(100 s^ -1 ) t + π/6] with the standard equation of SHM x = A (ω t + φ) , we get: Amplitude A = 2.0 cm Angular frequency ω = 100 rad/s Time period T = (2π)/(ω) = (2π)/(100) 0.063 s Spring constant k = mω^2 = (10 × 10^ -3 kg)(100 rad/s)^2 = 100 N/m

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 amplitude, the time period, and the spring constant for this motion.

Options

A $2.0 \text{ cm}$ , $0.063 \text{ s}$ , $10 \text{ N/m}$
B $2.0 \text{ cm}$ , $0.063 \text{ s}$ , $100 \text{ N/m}$
C $1.0 \text{ cm}$ , $0.063 \text{ s}$ , $50 \text{ N/m}$
D $4.0 \text{ cm}$ , $0.031 \text{ s}$ , $100 \text{ N/m}$

Correct answer

B. $2.0 \text{ cm}$ , $0.063 \text{ s}$ , $100 \text{ N/m}$

Step-by-step solution

Comparing the given equation

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

with the standard equation of SHM

x = A \sin(\omega t + \phi)

, we get:

Amplitude

A = 2.0 \text{ cm}

Angular frequency

\omega = 100 \text{ rad/s}

Time period

T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{100} \approx 0.063 \text{ s}

Spring constant

k = m\omega^2 = (10 \times 10^{-3} \text{ kg})(100 \text{ rad/s})^2 = 100 \text{ N/m}

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