Question

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 $\left( K =2 N m ^{-1}\right)$ If the block is horizontally displaced through ' $X$ ' $m$ then the number of complete oscillations it will make in $14 \pi$ seconds will be

Hint:

When two springs are connected in parallel, their spring constants add up to give an effective spring constant, and the time period depends on both the effective spring constant and the mass of the block.

Answer 1

Correct Answer: 20

The correct answer is 20.

Keff $=K+K$ as both springs are in use in parallel
$=2k$
$=2\times 2=4N/m$
$m=490gm$
$=0.49kg$
$T=2\pi \sqrt{\frac{m}{Keff}}=2\pi \sqrt{\frac{0.49kg}{4}}$
$=2\pi \sqrt{\frac{49}{400}}=2\pi \frac{7}{20}=\frac{7\pi }{10}$
No. of oscillation in the $14\pi$ is
$N=\frac{\text{time}}{T}=\frac{14\pi }{7\pi /10}=20$

Answer 2

The effective spring constant \( K_{\text{eff}} \) is the sum of the spring constants as both springs are in use in parallel: \[ K_{\text{eff}} = K + K = 2k = 4 \, \text{N/m} \] The mass of the block is: \[ m = 490 \, \text{g} = 0.49 \, \text{kg} \] The time period \( T \) of oscillation is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{K_{\text{eff}}}} \] Substituting the values: \[ T = 2\pi \sqrt{\frac{0.49}{4}} = 2\pi \sqrt{\frac{49}{400}} = 2\pi \times \frac{7}{20} = \frac{7\pi}{10} \, \text{seconds} \] The number of oscillations in 14t seconds is: \[ N = \frac{\text{time}}{T} = \frac{14t}{T} = \frac{14t}{7\pi/10} = 20 \] Thus, the number of complete oscillations is 20.