Question

A block of mass 1 kg is attached to one end of a spring of force constant k = 20 N/m. The other end of the spring is attached to a fixed rigid support. This spring block system is made to oscillate on a rough horizontal surface (µ = 0.04). The initial displacement of the block from the equilibrium position is a = 30 cm. How many times the block passes from the mean position before coming to rest? (Take g = 10 m/s²)

• A. 11
• B. 6
• C. 7
• D. 15

Answer 1

Answer: (C)

7

Answer 2

The kinetic friction force is calculated as $f_k = \mu N = \mu mg$. Given $m = 1$ kg, $g = 10$ m/s², and $\mu = 0.04$, the friction force is $f_k = 0.04 \times 1 \times 10 = 0.4$ N.

The amplitude of oscillation decreases by a constant amount $\Delta A$ in each half-oscillation, given by $\Delta A = \frac{2 f_k}{k}$. With $k = 20$ N/m, $\Delta A = \frac{2 \times 0.4}{20} = \frac{0.8}{20} = 0.04$ m.

The initial amplitude is $A_0 = 30$ cm $= 0.3$ m. The number of half-oscillations until the block comes to rest can be estimated by finding when the amplitude becomes less than or equal to $\Delta A$. The number of passes through the mean position is often approximated by $\lfloor \frac{A_0}{\Delta A} \rfloor$.

Number of passes $\approx \lfloor \frac{0.3 \text{ m}}{0.04 \text{ m}} \rfloor = \lfloor 7.5 \rfloor = 7$.

This means the block completes 7 half-oscillations and is in the process of the 8th half-oscillation when it stops. Each completed half-oscillation involves one pass through the mean position. Following the convention that the number of passes is $\lfloor A_0 / \Delta A \rfloor$, the block passes the mean position 7 times.