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For a particle executing SHM, x = displa...

For a particle executing SHM, x = displacement from mean position, v = velocity and a = acceleration at any instant, then

A

`v - x` graph is a circle

B

`v - x` graph is an ellipse

C

`a - x` graph is a straight line

D

`a - x` graph is a circle

Text Solution

Verified by Experts

The correct Answer is:
B, C
`v = omega sqrt (A^(2) - x^(2))`
`:. (v^(2))/(omega^(2)) + (x^(2))/((1)^(2)) = A^(2)`
i.e. `v - x` graph is an ellipse.
`a = - omega^(2)x`
i.e. `a - x` graph is a straight line passing through origin with negative slope.
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Knowledge Check

  • The velocity of particle in SHM at displacement y from mean position is

    A
    `omegasqrt((a^(2)+y^(2)))`
    B
    `omegasqrt((a^(2)-y^(2)))`
    C
    `wy`
    D
    `omega^(2)sqrt((a^(2)+y^(2)))`

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    A
    Infinity
    B
    Varies
    C
    Maximum
    D
    Zero

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    A
    20 N
    B
    15 N
    C
    `(20)/(3) N`
    D
    `(40)/(9) N`

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