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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

`(a)v - x` graph is a circle

B

`(b)v - x` graph is an ellipse

C

`(c)a - x` graph is a straight line

D

`(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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