For particle {\P} revolving round the centre {\O} with radius of circular path {\r} and angular velocity \omega, as shown in below figure, the projection of {\OP} on the {\x}-axis at time {\t} is [8-Apr-2023 shift 2]

Correct Answer is : x(t)=rcos(ωt+π6)

x(t)=r \cos (\omega t+\frac{\pi}{6})

x(t)=r \cos (\omega t-\frac{\pi}{6} \omega)

x(t)=r \sin (\omega t+\frac{\pi}{6})

JEE Mains 8-Apr-2023 Shift 2 Solved Paper

Section: Physics

Question : 47 of 90

Marks: +1, -0

For particle P revolving round the centre O with radius of circular path r and angular velocity ω, as shown in below figure, the projection of OP on the x-axis at time t is

[8-Apr-2023 shift 2]

x(t)=r‌cos(ωt+‌
π
6
)
x(t)=r‌cos(ωt−‌
π
6
ω)
x(t)=r‌cos(ωt)
x(t)=rsin‌(ωt+‌
π
6
)

Validate

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