A mass of Hg is executing SHM which is given by x=6.0 (100 t+(π)/(4)) cm . What is the maximum kinetic energy?

Correct answer: D. 18 J Here, m=1 ~kg The given equation of SHM is x=6.0 (100 t+(π)/(4)) Comparing it with equation of SHM, aligned x & =A (ω t+φ), we have A & =6.0 ~cm=(6)/(100) ~m aligned and ω=100 rad / s Maximum kinetic energy =(1)/(2) ~m(v_ )^2 =(1)/(2) ~m(~A ω)^2=(1)/(2) × 1 ×[(6)/(100) × 100]^2=18 ~J

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Oscillations

JIPMER 2018 · Physics · Oscillations

A mass of Hg is executing SHM which is given by x=6.0 (100...

A mass of Hg is executing SHM which is given by

x=6.0 \cos \left(100 t+\frac{\pi}{4}\right) \mathrm{cm}

. What is the maximum kinetic energy?

Options

A $3 J$
B $6 J$
C $9 J$
D $18 J$

Correct answer

D. $18 J$

Step-by-step solution

Here,

\mathrm{m}=1 \mathrm{~kg}

The given equation of SHM is

x=6.0 \cos \left(100 t+\frac{\pi}{4}\right)

Comparing it with equation of SHM,

\begin{aligned} x & =A \cos (\omega t+\phi), \text{ we have } \\ A & =6.0 \mathrm{~cm}=\frac{6}{100} \mathrm{~m}\end{aligned}

and

\quad \omega=100 \mathrm{rad} / \mathrm{s}

Maximum kinetic energy

=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\max }\right)^2

=\frac{1}{2} \mathrm{~m}(\mathrm{~A} \omega)^2=\frac{1}{2} \times 1 \times\left[\frac{6}{100} \times 100\right]^2=18 \mathrm{~J}

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