Oscillations Question 10
Question 10 - 31 January - Shift 1
In the figure given below. a block of mass $M=490 g$ placed on a frictionless table is connected with two springs having same spring constant $(K=2 N m^{-1})$. If the block is horizontally displaced through ’ $X$ ’m then the number of complete oscillations it will make in $14 \pi$ seconds will be
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Answer: (20)
Solution:
Formula: Time period of Simple harmonic motion, Combination of springs
Keff $=K+K$ as both springs are in use in parallel
$=2 k$
$=2 \times 2=4 N / m$
$ \begin{aligned} m & =490 gm \\ & =0.49 kg \end{aligned} $
$T=2 \pi \sqrt{\frac{m}{Keff}}=2 \pi \sqrt{\frac{0.49 kg}{4}}$
$ =2 \pi \sqrt{\frac{49}{400}}=2 \pi \frac{7}{20}=\frac{7 \pi}{10} $
No. of oscillation in the $14 \pi$ is
$ N=\frac{\text{ time }}{T}=\frac{14 \pi}{7 \pi / 10}=20 $
Ans in 20.
