A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.

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Solution

Time period of simple pendulum

$= 2 π \sqrt{(\frac{l}{g})}$

Time period of spring is $2 π \sqrt{(\frac{m}{k})}$

$T_{p} = T_{s}$ [Frequency is small ]

$\Rightarrow \sqrt{(\frac{l}{g})} = \sqrt{(\frac{m}{k})}$

$\Rightarrow (\frac{l}{g}) = (\frac{m}{k})$

$\Rightarrow l = \frac{m g}{k} = \frac{F}{k} = x$

(Because, restoring force = weight = F = mg)

$\Rightarrow l = x$ (Proved)

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A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.

Time period of simple pendulum =2π √(lg) Time period of spring is 2π √(mk) Tp=Ts [Frequency is small ] ⇒ √(lg)=√(mk) ⇒ (lg)=(mk) ⇒ l=mgk=Fk=x (Because, restoring force = weight = F = mg) ⇒ l=x (Proved)