placed on a frictionless horizontal floor are connected with the help of a spring of force constant
.The discs are also connected with two light rods each of length
that are pivoted to a nail driven into the floor as shown in the figure by a top view.
If period of small oscillations of the system is
Find relaxed length of spring.
Answer comes in the form of
Type
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Consider the origin of the X-Y plane to be placed at the pivot. The X-axis is horizontal to the right and the Y-axis points upwards. At any instant, let each of the rods make an angle θ with the horizontal. Therefore, the coordinates of each mass is:
x 1 = L cos θ y 1 = L sin θ
x 2 = L cos θ y 2 = − L sin θ
The kinetic energy of the system is:
T = m L 2 θ ˙ 2
The spring potential energy of the system is:
V = 2 k ( 2 L sin θ − L o ) 2
Where L o is the natural length of the spring. Lagrange's equation yields the following equation of motion. This can also be easily derived by computing the net torque experienced by a mass about the pivot. This working is left out of the solution. The equation of motion is:
2 m L 2 θ ¨ + 2 L k ( 2 L sin θ − L o ) cos θ = 0
Now, consider the system in equilibrium. Say the angle the rods make with the horizontal is θ o . Then, by using some elementary trigonometry, it can be seen that:
2 L sin θ = L o
Since we are considering small oscillations about this equilibrium, consider:
θ = ϕ + θ o
Here, ϕ is small enough so that the small-angle approximation holds good. That means:
sin ϕ ≈ ϕ cos ϕ ≈ 1
Plugging the above substitution in the EOM and simplifying gives:
m L 2 ϕ ¨ + ( 2 k L 2 cos 2 θ o ) ϕ = 0
The above equation represents SHM with natural frequency:
ω 2 = M 2 k cos 2 θ o
It is already given that:
ω 2 = M k
This implies that:
2 cos 2 θ o = 1
We know that
sin θ o = 2 L L o
⟹ 2 ( 1 − 4 L 2 L o 2 ) = 1 L o = 2 L