A particle of mass m is attached to 3 springs | vedclass

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(A) Let the particle be displaced by a small distance $x$ along the direction of spring $C$. The displacement $x$ can be resolved into components alon

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{m}{{2k}}} $

Vedclass · Answer

(A) Let the particle be displaced by a small distance $x$ along the direction of spring $C$. The displacement $x$ can be resolved into components along the axes of the three springs. Spring $C$ undergoes a compression of $x$. Spring $A$ and $B$ are at $45^{\circ}$ and $90^{\circ}$ respectively to the direction of displacement. The components of displacement along springs $A$ and $B$ are $x \cos(45^{\circ}) = \frac{x}{\sqrt{2}}$ and $x \cos(45^{\circ}) = \frac{x}{\sqrt{2}}$ respectively. The restoring forces are $F_C = Kx$,$F_A = K \frac{x}{\sqrt{2}}$,and $F_B = K \frac{x}{\sqrt{2}}$. The total restoring force $F_{net}$ along the direction of displacement is the sum of the components of these forces along that direction: $F_{net} = F_C + F_A \cos(45^{\circ}) + F_B \cos(45^{\circ}) = Kx + (K \frac{x}{\sqrt{2}}) \frac{1}{\sqrt{2}} + (K \frac{x}{\sqrt{2}}) \frac{1}{\sqrt{2}} = Kx + \frac{Kx}{2} + \frac{Kx}{2} = 2Kx$. Comparing this with $F = k_{eff} x$,we get $k_{eff} = 2K$. The time period is $T = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{m}{2K}}$.

$A$ particle of mass $m$ is attached to $3$ springs $A$,$B$ and $C$ of equal force constant $K$ as shown in the figure. If the particle is pushed slightly along the line of spring $C$ and released,find the period of oscillation.

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