A body of mass 1 kg is executing simple harmonic motion. Its displacement y (cm) at t seconds is given by y = 6 (100t + (π)/(4)). Its maximum kinetic energy is

Correct answer: B. 18J Given, a = 6cm = 6 × 10 ^ - 2 m ω = 100rads^ - 1 , m = 1kg K _ = (1)/(2)m ω ^ 2 a^ 2 = (1)/(2) × 1 × (100)^ 2 × (6 × 10 ^ - 2 )^ 2 = 18J

BITSAT 2017

Exam: BITSAT
Year: 2017
Subject: Physics
Chapter: Oscillations

BITSAT 2017 · Physics · Oscillations

A body of mass 1 kg is executing simple harmonic motion. Its...

A body of mass $1$ kg is executing simple harmonic motion. Its displacement $y \left(cm\right)$ at $t$ seconds is given by $y = 6\sin \left(100t +$ $\dfrac{\pi}{4}\right).$ Its maximum kinetic energy is

Options

A $6J$
B $18J$
C $24J$
D $36J$

Correct answer

B. $18J$

Step-by-step solution

Given, $a = 6cm = 6 \times {10}^{- 2}m$

$\omega = 100rads^{- 1}, m = 1kg$

${K}_\max = \dfrac{1}{2}m{ω}^{2}a^{2}$

$= \dfrac{1}{2} \times 1 \times \left(100\right)^{2} \times \left(6 \times {10}^{- 2}\right)^{2}$

$= 18J$

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