The 2 spring problem

Classical Mechanics Level 3

A uniform rod of mass $M$ and length $L$ is pivoted at the center. Its two ends are attached to two springs of spring constants $k$ . The springs are fixed to the rigid support as shown in the figure , and the rod is free to rotate in the horizontal plane.The rod is gently pushed through small angle $\theta$ in one direction and then released. What is the frequency of oscillation?

\frac{1}{2\pi}\sqrt{\frac{6k}{M}}

\frac{1}{2\pi}\sqrt{\frac{2k}{M}}

\frac{1}{2\pi}\sqrt{\frac{k}{M}}

\frac{1}{2\pi}\sqrt{\frac{24k}{M}}

1 solution

Md Zuhair
Jul 25, 2016

Relevant wiki: Simple Harmonic Motion - Problem Solving

Ok so this a an $\text{IITJEE problem (2009}$ ...

Hence we begin to solve the problem ... Here we are getting two springs... we will be taking the torque about the centre in this Rod ... Now we proceed:-

So We may write the equation as ,

$\implies kx.\dfrac{L}{2} + kx .\dfrac{L}{2} = I\alpha$

Now here $x = \dfrac{L}{2}\theta$ Hence :-

$\implies k \dfrac{L^2 \theta}{4} + k \dfrac{L^2 \theta}{4} = \dfrac{mL^2 \alpha}{12}$

$\implies \dfrac{k L^2 \theta}{2} =\dfrac{mL^2 \alpha}{12}$

Futher solving we will get

$\dfrac{\alpha}{θ} = \dfrac{6k}{m}$ .... And $T=2\pi \sqrt{\dfrac{θ}{a}}$

Hence we get $\dfrac{1}{T} =freq = \dfrac{1}{2 \pi} \sqrt{\dfrac{6k}{m}}$

Where did you get the 12 x?shoulnt it be ml^2/4.a?

Mr Yovan - 4 years, 10 months ago

Hey.. Its a Rod .... Hence I = ML^2/12 ... Is that what u r asking for?

Md Zuhair - 4 years, 10 months ago

The 2 spring problem

1 solution

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