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  4. A particle is executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 and its acceleration is 0.5 m / s 2. The maximum velocity of the particle is ( in m / s )

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Q. A particle is executing simple harmonic motion with amplitude of $0.1\, m$. At a certain instant when its displacement is $0.02$ and its acceleration is $0.5\, m / s ^{2}$. The maximum velocity of the particle is $( in m / s )$

ManipalManipal 2011Oscillations

Solution:

Maximum acceleration
$\alpha=\omega^{2} y$
$\therefore 0.5=\omega^{2} \times 0.02$
or $\omega^{2}=\frac{0.5}{0.02}=25$
So, $\omega=5$
Now, maximum velocity is
$v=a \omega=0.1 \times 5=0.5\, m / s$