Mass ' $m$ ' is suspended from the mass-less pulley which itself is supported by two springs connected to each other as shown. If the mass is slightly displaced from its equilibrium position, find the frequency of resulting SHM of mass ' $m$ '.

f = \frac{1}{2 π} \sqrt{\frac{4 k_{1} k_{2}}{(k_{1} + k_{2}) m}}

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f = \frac{1}{2 π} \sqrt{\frac{k_{1} k_{2}}{(k_{1} + k_{2}) m}}

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f = \frac{1}{2 π} \sqrt{\frac{k_{1} k_{2}}{4 (k_{1} + k_{2}) m}}

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f = \frac{1}{2 π} \sqrt{\frac{4 (k_{1} + k_{2})}{m}}

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f = \frac{1}{2 π} \sqrt{\frac{(k_{1} + k_{2})}{m}}

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Solution

The correct option is B $f = \frac{1}{2 π} \sqrt{\frac{4 k_{1} k_{2}}{(k_{1} + k_{2}) m}}$

Let mass $m$ be displaced by a distance of $x$ from its equilibrium position
so that the additional elongation in spring 1 & 2 are $x_{1}$ & $x_{2}$ respectively, then
$2 x = x_{1} + x_{2}$ -----(1)
Also since tension in both springs are same.
$T = k_{1} x_{1} = k_{2} x_{2}$
from (1) we have
$2 x = x_{1} (1 + \frac{k_{1}}{k_{2}})$
$∴ T = k_{1} x_{1} = \frac{2 k_{1} x}{1 + \frac{k_{1}}{k_{2}}} = \frac{2 k_{1} k_{2} x}{k_{1} + k_{2}}$
$T^{'} = 2 T = \frac{4 k_{1} k_{2} x}{k_{1} + k_{2}}$
$T$ and $T^{'}$ are additional tension, over and above the tension already present at equilibrium. hence, for the mass $m$
$- \frac{m d^{2} x}{d t^{2}} = \frac{4 k_{1} k_{2} x}{k_{1} + k_{2}}$
$f = \frac{1}{2 π} \sqrt{\frac{4 k_{1} k_{2}}{(k_{1} + k_{2}) m}}$

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Mass 'm' is suspended from the mass-less pulley which itself is supported by two springs connected to each other as shown. If the mass is slightly displaced from its equilibrium position, find the frequency of resulting SHM of mass 'm'.

Mass ' $m$ ' is suspended from the mass-less pulley which itself is supported by two springs connected to each other as shown. If the mass is slightly displaced from its equilibrium position, find the frequency of resulting SHM of mass ' $m$ '.