Springs of spring constants K, 2K, 4K, 8K, do | vedclass

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(A) The equivalent spring constant $K_{eq}$ for springs in series is given by $\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} + \dot

Vedclass · Accepted Answer

$0.13$

Vedclass · Answer

(A) The equivalent spring constant $K_{eq}$ for springs in series is given by $\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} + \frac{1}{K_3} + \dots$Substituting the given values: $\frac{1}{K_{eq}} = \frac{1}{K} + \frac{1}{2K} + \frac{1}{4K} + \frac{1}{8K} + \dots$Factoring out $\frac{1}{K}$: $\frac{1}{K_{eq}} = \frac{1}{K} [1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots]$The term in the bracket is an infinite geometric progression with first term $a = 1$ and common ratio $r = 1/2$. The sum is $S = \frac{a}{1-r} = \frac{1}{1 - 1/2} = 2$.Thus,$\frac{1}{K_{eq}} = \frac{1}{K} 	imes 2 = \frac{2}{K}$,which implies $K_{eq} = \frac{K}{2}$.Given $K = 2 \, N/cm = 200 \, N/m$,so $K_{eq} = \frac{200}{2} = 100 \, N/m$.The mass $m = 40 \, g = 0.04 \, kg$.The time period $T = 2 \pi \sqrt{\frac{m}{K_{eq}}} = 2 \pi \sqrt{\frac{0.04}{100}} = 2 \pi \sqrt{0.0004} = 2 \pi 	imes 0.02 = 0.04 \pi \approx 0.04 	imes 3.14 = 0.1256 \approx 0.13 \, s$.

Springs of spring constants $K, 2K, 4K, 8K, \dots$ are connected in series. $A$ mass of $40 \, g$ is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillation in seconds? (Given $K = 2 \, N/cm$)

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