My Mechanics Problem

Classical Mechanics Level 4

Consider a thin rod having mass m m and of length l l pivoted at one of its end. It is held by a spring at its midpoint and another at its far end, both pulling in opposite directions. The springs have spring constant k k , and at equilibrium their pull is perpendicular to the rod. If (for small oscillations)- ω 2 = a k b m − c g d l \omega^2 = \frac{ak}{bm} - \frac{cg}{dl}

Find a + c b + d \frac{a+c}{b+d}


The answer is 3.

Vatsalya Tandon by Vatsalya Tandon , 21, India

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1 solution

Parth Sankhe
Oct 19, 2018

Torque = k ( 0.5 l θ ) 0.5 l + k ( l θ ) l + m g ( 0.5 l ) θ k(0.5l\theta)0.5l + k(l\theta)l + mg(0.5l)\theta

= ( 5 4 k l 2 + m g l 2 ) θ (\frac {5}{4}kl^2 + mg\frac {l}{2})\theta

= c θ c\theta

w 2 = c I = c ⅓ m l 2 = 15 k 4 m + 3 g 2 l w^2=\frac {c}{I}=\frac {c}{⅓ml^2}= \frac {15k}{4m} + \frac {3g}{2l}

Thus, the answer = 15 + 3 4 + 2 = 3 \frac {15+3}{4+2}=3

Note: This solution heavily relies on certain calculation mechanics which are greatly used in SHM calculations, I suggest reading those first.

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