Question 174: Oscillations

Question

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 _____________.

Accepted Answer

$\mathrm{K_eff}=\mathrm{K}+\mathrm{K}$ $=2 \mathrm{k}$ (As both springs are in use in parallel)
$=2 	imes 2=4 \mathrm{~N} / \mathrm{m}$
$$
\mathrm{m}=490 \mathrm{gm}
$$
$$
\begin{aligned}
& =0.49 \mathrm{~kg}
\end{aligned}
$$
$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{Keff}}}=2 \pi \sqrt{\frac{0.49 \mathrm{~kg}}{4}}$
$$
=2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10}
$$
No. of oscillation in the $14 \pi$ is
$$
\mathrm{N}=\frac{	ext { time }}{\mathrm{T}}=\frac{14 \pi}{7 \pi / 10}=20
$$

Oscillations

Integer / Numeric Answer Type