KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 25 of 120

Marks: +1, -0

One end of a rod of length L is fixed to a point on the circumference of a wheel of radius R. The other end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure below. The wheel is rotating with aconstant angular velocity ω about O. Taking

T ={2π}/ω

the motion of the rod is

simple harmonic with a period of T
simple harmonic with a period of T/2
not simple harmonic but periodic with a period of T
not simple harmonic but periodic with a period of T/2

Validate

Solution:

\cos θ = {R ^2+x^2- L^2}/{2Rx}

⇒

x^ 2 = 2Rx \cosecθ + L^2 -R^ 2

displacement of S.H. M. is in the form of

x = A \sin ωt + c

Therefore it is not S. H. M.
It is S. H. M. only which.
It L = R.
⇒

x^2 = 2Rx \cos θ

⇒

x = 2R \cos θ

x = 2R cos ωt [ S.H. M.]
But period of this motion is T

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