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The speed v of a particle moving along a...

The speed v of a particle moving along a straight line. When it is at distance x from a fixed point on the line is `v^(2)=144-9x^(2)`. Select the correct alternatives

A

The magnitude of acceleration at a distance 3 units from the fixed point is 27 units

B

the motion is simple harmonic with `T=(2pi)/(3)` units

C

The maximum displacement from the fixed point is 4 units

D

all are correct

Text Solution

Verified by Experts

The correct Answer is:
D
(d) `v=omegasqrt(A^(2)+x^(2))" "`(in SHM)
or `v^(2)=omega^(2)A^(2)-omega^(2)x^(2)`
Comparing the given equation with this equation we get,
`omega^(2)=9`
`therefore omega=3=(2pi)/(T)`
`therefore T=(2pi)/(3)` units
Also , `omega^(2)A^(2)=144`
`therefore A=4 "units", |a|=omega^(2)x=(9)(3)=27` units
Displacement ` le` distance
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