### 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

## Show Answer

#### 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.