In the figure given below, a block of mass placed on a frictionless table is connected with two springs having the same spring constant If the block is horizontally displaced through then the number of complete oscillations it will make in seconds wil

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Solution

Q.
In the figure given below, a block of mass $M = 490 g$ placed on a frictionless table is connected with two springs having the same spring constant $(K = 2 {Nm}^{- 1}) .$ If the block is horizontally displaced through $' X' m,$ then the number of complete oscillations it will make in $14 π$ seconds will be ______ . [2023]

Ans.
(20)

$K_{eff} = K + K (as both springs are in use in parallel) = 2 k = 2 \times 2 = 4 N/m$

$m = 490 gm = 0.49 kg$

$T = 2 π \sqrt{\frac{m}{K_{eff}}} = 2 π \sqrt{\frac{0.49}{4}}$

$= 2 π \sqrt{\frac{49}{400}} = 2 π \frac{7}{20} = \frac{7 π}{10}$

$Number of oscillations in 14 π is$

$N = \frac{time}{T} = \frac{14 π}{7 π / 10} = 20$

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