A tray of mass 12 ,kg is supported by two ide | vedclass

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(C) Mass of the tray, $m = 12 \,kg$. Time period, $T = 1.5 \,s$. Let $k$ be the spring constant of each spring. Since the springs are connected in par

Vedclass · Accepted Answer

$105 \,Nm^{-1}$

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(C) Mass of the tray, $m = 12 \,kg$. Time period, $T = 1.5 \,s$. Let $k$ be the spring constant of each spring. Since the springs are connected in parallel, the effective spring constant is $k_{	ext{net}} = k + k = 2k$. The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k_{	ext{net}}}}$. Substituting the values, we get $1.5 = 2\pi \sqrt{\frac{12}{2k}}$. $1.5 = 2\pi \sqrt{\frac{6}{k}}$. Squaring both sides, we have $(1.5)^2 = (2\pi)^2 \cdot \frac{6}{k}$. $2.25 = 4\pi^2 \cdot \frac{6}{k}$. $k = \frac{24\pi^2}{2.25} \approx \frac{24 	imes 9.87}{2.25} \approx 105.28 \,Nm^{-1}$. Thus, the spring constant of each spring is approximately $105 \,Nm^{-1}$.

$A$ tray of mass $12 \,kg$ is supported by two identical springs as shown in the figure. When the tray is pressed down slightly and then released, it executes $SHM$ with a time period of $1.5 \,s$. The spring constant of each spring is

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