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Question

A particle is placed at the lowest point of a smooth wire frame in the shape of a parabola, lying in the vertical xy plane having equation x2=5y, where (x,y) are in metres. After slight displacement, the particle is set free to move. Find the angular frequency of oscillation in rad/s.
[Take g=10 m/s2)]

A
2 rad/s
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B
4 rad/s
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C
6 rad/s
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D
8 rad/s
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Solution

The correct option is A 2 rad/s


Given, y=x25
Differentiating with respect to x on both sides we get,

dydx=0.4x

On drawing the tangent at the position of particle, we get tanθ as the slope.
slope=tanθ=dydx , at the given position on the wire, we get tanθ=0.4x

The value of θ will be very small, as the particle is given a slight displacement from the lowest position.
tanθsinθθ
Or, we can say that, sinθ0.4x

From figure we can deduce that, direction of net restoring force (mgsinθ) is opposite to the direction of displacement of particle.
Hence,
Frestoring=mgsinθ
ma=mgsinθ
a=10(0.4x)=4x
Comparing this with a=ω2x, we have
ω2=4ω=2 rad/s

Thus, option (a) is the correct answer.

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