Small Oscillation

Classical Mechanics Level pending

Two identical small discs each of mass m m placed on a frictionless horizontal floor are connected with the help of a spring of force constant k k .The discs are also connected with two light rods each of length l l 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 2 π m k 2 \pi \sqrt{\frac{m}{k}}

Find relaxed length of spring.
Answer comes in the form of α l \alpha l
Type α \alpha


The answer is 1.414.

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1 solution

Karan Chatrath
Sep 27, 2020

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 \theta with the horizontal. Therefore, the coordinates of each mass is:

x 1 = L cos θ x_1= L\cos{\theta} y 1 = L sin θ y_1 = L\sin{\theta}

x 2 = L cos θ x_2= L\cos{\theta} y 2 = L sin θ y_2 = -L\sin{\theta}

The kinetic energy of the system is:

T = m L 2 θ ˙ 2 \mathcal{T} = mL^2\dot{\theta}^2

The spring potential energy of the system is:

V = k 2 ( 2 L sin θ L o ) 2 \mathcal{V} = \frac{k}{2}(2L\sin{\theta} - L_o)^2

Where L o 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 2mL^2\ddot{\theta} + 2Lk(2L\sin{\theta}-L_o)\cos{\theta} =0

Now, consider the system in equilibrium. Say the angle the rods make with the horizontal is θ o \theta_o . Then, by using some elementary trigonometry, it can be seen that:

2 L sin θ = L o 2L\sin{\theta} = L_o

Since we are considering small oscillations about this equilibrium, consider:

θ = ϕ + θ o \theta = \phi + \theta_o

Here, ϕ \phi is small enough so that the small-angle approximation holds good. That means:

sin ϕ ϕ \sin{\phi}\approx \phi cos ϕ 1 \cos{\phi} \approx 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 mL^2\ddot{\phi} + (2kL^2\cos^2{\theta_o})\phi=0

The above equation represents SHM with natural frequency:

ω 2 = 2 k cos 2 θ o M \omega^2 = \frac{2k\cos^2{\theta_o}}{M}

It is already given that:

ω 2 = k M \omega^2 = \frac{k}{M}

This implies that:

2 cos 2 θ o = 1 2\cos^2{\theta_o} = 1

We know that

sin θ o = L o 2 L \sin{\theta_o} = \frac{L_o}{2L}

2 ( 1 L o 2 4 L 2 ) = 1 \implies 2\left(1 - \frac{L_o^2}{4L^2}\right) = 1 L o = 2 L L_o = \sqrt{2}L

@Karan Chatrath Thanks for the solution

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath can you please help me in this problem

Talulah Riley - 8 months, 1 week ago

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Is the answer 5?

Karan Chatrath - 8 months, 1 week ago

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@Karan Chatrath After 5 hours I just checked brilliant now. And you replied now only.

Talulah Riley - 8 months, 1 week ago

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@Talulah Riley I edited my comment. Anyway, do let me know.

Karan Chatrath - 8 months, 1 week ago

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@Karan Chatrath @Karan Chatrath He was also online on whatsApp. He replied, he doesn't know the answer.
Sir can you share your approach. Thanks in advance.

Talulah Riley - 8 months, 1 week ago

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@Talulah Riley Show your attempt. I'll comment on that

Karan Chatrath - 8 months, 1 week ago

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@Karan Chatrath @Karan Chatrath Ok , I will come online on Brilliant at 10:30 PM sharp with my approach. BYE

Talulah Riley - 8 months, 1 week ago

@Karan Chatrath This question was asked by my friend. So let me ask him and will clarify you.

Talulah Riley - 8 months, 1 week ago

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