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Statement (a) : The velocity of simple harmonic oscillator is maximum at mean position

Statement (b) : At extreme position the acceleration of simple harmonic oscillator is maximum.

Statement (c) : The velocity of simple harmonic oscillator is minimum at extreme position

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all are true

only a is true

a, b are true

b, c are true

answer is D.

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Detailed Solution

If V be the magnitude of velocity of a particle when its distance from equilibrium position is x, then $\large v\, = \,\omega \sqrt {{A^2} - {x^2}}$

Also at that instant acceleration $\large a\, = \, - {\omega ^2}x$ where $\large \omega$ is the angular velocity.

Clearly V is maximum when x = 0 and minimum when x = A. Again 'a' is maximum when x is maximum is when x = A so (4) is right option.

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