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(B) When the block is displaced vertically downwards by a small distance $x$,the central spring is stretched by $x$,providing a restoring force $F_1 =

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

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(B) When the block is displaced vertically downwards by a small distance $x$,the central spring is stretched by $x$,providing a restoring force $F_1 = kx$ upwards.Each of the two side springs,which are at an angle of $45^{\circ}$ to the horizontal,will also be stretched by an amount $\Delta l = x \sin 45^{\circ} = \frac{x}{\sqrt{2}}$.The restoring force in each side spring is $F_s = k \Delta l = \frac{kx}{\sqrt{2}}$.The vertical component of the force from each side spring is $F_v = F_s \sin 45^{\circ} = \left( \frac{kx}{\sqrt{2}} ight) \left( \frac{1}{\sqrt{2}} ight) = \frac{kx}{2}$.The total restoring force $F_{net}$ acting on the block is the sum of the forces from all three springs:$F_{net} = F_1 + 2 	imes F_v = kx + 2 	imes \left( \frac{kx}{2} ight) = kx + kx = 2kx$.Since $F_{net} = ma$,we have $ma = 2kx$,which gives $a = \left( \frac{2k}{m} ight) x$.Comparing this with the standard $SHM$ equation $a = \omega^2 x$,we get $\omega^2 = \frac{2k}{m}$,so $\omega = \sqrt{\frac{2k}{m}}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{2k}}$.

Initially,the system is in equilibrium. The time period of $SHM$ of the block in the vertical direction is

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