JEE Advance - Physics (2022 - Paper 2 Online - No. 7)
On a frictionless horizontal plane, a bob of mass $m=0.1 \mathrm{~kg}$ is attached to a spring with natural length $l_{0}=0.1 \mathrm{~m}$. The spring constant is $k_{1}=0.009 \,\mathrm{Nm}^{-1}$ when the length of the spring $l>l_{0}$ and is $k_{2}=0.016 \,\mathrm{Nm}^{-1}$ when $l < l_{0}$. Initially the bob is released from $l=$ $0.15 \mathrm{~m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) s$, then the integer closest to $n$ is __________.
Answer
6
Explanation
Time period of oscillation
$$ \begin{aligned} & =\mathrm{T}_1+\mathrm{T}_2 \\\\ & =\pi \sqrt{\frac{m}{K_1}}+\pi \sqrt{\frac{m}{K_2}} \\\\ & =\pi \sqrt{\frac{0.1}{0.009}}+\pi \sqrt{\frac{0.1}{0.016}} \\\\ & =\frac{\pi}{0.3}+\frac{\pi}{0.4} \\\\ & =\frac{\pi(0.4+0.3)}{0.12} \\\\ & =\frac{70 \pi}{12} \\\\ & =5.83 \pi \text { seconds } \\\\ & \simeq 6 \pi \text { seconds } \end{aligned} $$
$$ \begin{aligned} & =\mathrm{T}_1+\mathrm{T}_2 \\\\ & =\pi \sqrt{\frac{m}{K_1}}+\pi \sqrt{\frac{m}{K_2}} \\\\ & =\pi \sqrt{\frac{0.1}{0.009}}+\pi \sqrt{\frac{0.1}{0.016}} \\\\ & =\frac{\pi}{0.3}+\frac{\pi}{0.4} \\\\ & =\frac{\pi(0.4+0.3)}{0.12} \\\\ & =\frac{70 \pi}{12} \\\\ & =5.83 \pi \text { seconds } \\\\ & \simeq 6 \pi \text { seconds } \end{aligned} $$
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