IIT JEE , AIEEE Forum - Physics , Mechanics

pinnaka

That is a simple Question
Here u need to consider the torques exerted by two different forces
SO when the Stick or teh rod makes an angle theta
Then torques due to two forces act on the rod
One is due to garvity and the other is due to the restoring force exerted by the spring and by the earth

NOw lets us consider the situation when the stick has made an angle theta with the vertical. Then we have the torque due to gravity to be
tau_g = -mgsin( theta)*l/2
the force due to garvity acts through the center of mass therfore the lever arm is l/2 sin(theta)
Now the torque due to the restoring force exerted by the spring is equal to -kx sin(90-theta) *l = tau_spring
the angle is deduced from geometry
The torques due to forces point in the same direction. therefore they can be added them up
tau_net = -mgsin(theta)*l/2- kxcos(theta)*l
where l is the lenght of the rod which should have been given in the question
for very small angles we have cos(theta) = 1
sin (theta) = theta
these results follows from talyor series
tau_net = -mg(theta)*l/2 - kx*l
now we have theta = x/l so x =theta*l
tau_net = - (mg*l/2-kl^2) theta
now we have the torque acting as a function of the angle rotated
therefore omega = sqrt{ ( mg*l/2+kl^2)/(1/3 ml^2)}= sqrt{3 mg/2l + 6k/2m}
the rod rotates about the pivot
T = 2 pie /omega
which gives u
T = 2 pie/omega
Now lets us test this result when u r spring constant is equal to zero
which means that no spring is present
then u have T to be a expected result
therfore the preiod
T = 2(pi) * sqrt{2ml/(6Kl+3mg}

----Hope that Helps

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