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Standard XIII

Physics

SHM expression

A particle ís...

Question

A particle ís executing simple harmonic motion with amplitude of $0.1 m$ . At a certain instant when its displacement is $0.02 m$ , its acceleration is $0.5 {ms}^{- 2}$ . The maximum velocity of the particle is (in ${ms}^{- 1}$ )

0.01

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0.05

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0.5

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0.25

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Solution

The correct option is C $0.5$
Acceleration, $a = 0.5 {ms}^{- 2}$
Amplitude, $A = 0.1 m$
Displacement, $y = 0.02 m$
Using the formula of maximum acceleration $a = ω^{2} y$
or $0.5 = ω^{2} \times 0.02$
or $ω^{2} = \frac{0.5}{0.02} = 25$
So, $ω = 5$
Now, maximum velocity is
$v = A ω = 0.1 \times 5 = 0.5 {ms}^{- 1}$

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SHM expression

Standard XIII Physics

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A particle ís executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 ms−2. The maximum velocity of the particle is (in ms−1)

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