A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in $m s e c^{- 1}$ at the centre of oscillation is

[JIPMER 1997]

20π

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100

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40π

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100π

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Solution

The correct option is C
40π

At centre $v_{m a x} = a ω = a \frac{2 π}{T} = \frac{0.2 \times 2 π}{0.01} = 40 π$

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A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in m sec−1 at the centre of oscillation is [JIPMER 1997]

The correct option is C 40π At centre vmax=aω=a2πT=0.2×2π0.01=40π

A simple harmonic oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The magnitude of the velocity in $m s e c^{- 1}$ at the centre of oscillation is

[JIPMER 1997]

The correct option is C
40π

At centre $v_{m a x} = a ω = a \frac{2 π}{T} = \frac{0.2 \times 2 π}{0.01} = 40 π$