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If s=a sinomegat hati+b cos omegat hatj,...

If s=a `sinomegat hati+b cos omegat hatj`, the equation of path of particle is

A

`x^(2)+y^(2)=sqrt(a^(2)+b^(2))`

B

`(x^(2))/(b^(2))+(y^(2))/(a^(2))=1`

C

`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
Here, x= a sin `omegat`
`therefore" "sin omegat=(x)/(a)`
Similarly, `y=b cos omega t`
`therefore" "cos omega t=(y)/(b)`
`therefore" "sin^(2)omegat+cos^(2)omegat=(x^(2))/(a^(2))+(y^(2))/(b^(2))`
`therefore" "(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` (Ellipse)
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Knowledge Check

  • The position vector of a particle is r = a sin omega t hati +a cos omega t hatj The velocity of the particle is

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  • Position vectors of a particle moving in xy plane at time t is vecr=a(1-cos ometat)hati+asin omegat hatj . The path of the particle is

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