Question 25

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 ______.

Correct Answer: 20

$$F_{\text{net}} = F_1 + F_2 = -Kx - Kx = -2Kx$$

Hence, $$K_{\text{eq}} = 2K = 2 \times 2 = 4\text{ N m}^{-1}$$

$$T = 2\pi \sqrt{\frac{0.49}{4}}$$

$$T = 2\pi \times \frac{0.7}{2} = 0.7\pi\text{ seconds}$$

Number of oscillations, $$N = \frac{t}{T} = \frac{14\pi}{0.7\pi}$$

$$N = \frac{14}{0.7} = \frac{140}{7} = 20$$

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