JEE MAIN - Physics (2023 - 31st January Morning Shift - No. 20)
In the figure given below, a block of mass $$M=490 \mathrm{~g}$$ placed on a frictionless table is connected with two springs having same spring constant $$\left(\mathrm{K}=2 \mathrm{~N} \mathrm{~m}^{-1}\right)$$. If the block is horizontally displaced through '$$\mathrm{X}$$' $$\mathrm{m}$$ then the number of complete oscillations it will make in $$14 \pi$$ seconds will be _____________.
_31st_January_Morning_Shift_en_20_1.png)
Answer
20
Explanation
$\mathrm{K_eff}=\mathrm{K}+\mathrm{K}$ $=2 \mathrm{k}$ (As both springs are in use in parallel)
$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
$$ \mathrm{m}=490 \mathrm{gm} $$
$$ \begin{aligned} & =0.49 \mathrm{~kg} \end{aligned} $$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $$
No. of oscillation in the $14 \pi$ is
$$ \mathrm{N}=\frac{\text { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20 $$
$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
$$ \mathrm{m}=490 \mathrm{gm} $$
$$ \begin{aligned} & =0.49 \mathrm{~kg} \end{aligned} $$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $$
No. of oscillation in the $14 \pi$ is
$$ \mathrm{N}=\frac{\text { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20 $$
Comments (0)
