Solution to Question 10
1. System Parameters
We are given a block of mass $m = 1.0$ kg attached to a spring. The block is on a rough floor.
- Spring constant $k = 5.0$ N/cm $= 500$ N/m.
- Coefficient of friction $\mu = 0.25$.
- Frictional Force $f = \mu m g = 0.25 \times 1.0 \times 10 = 2.5$ N.
2. Derivation of Amplitude Decay
Work-Energy Analysis for one Half-Cycle:
Consider the block moving from an extreme position $x = A_n$ to the opposite extreme position $x = -A_{n+1}$. The velocity is zero at both points.
Initial Mechanical Energy: $E_i = \frac{1}{2} k A_n^2$
Final Mechanical Energy: $E_f = \frac{1}{2} k A_{n+1}^2$
Work done by Friction: The friction force $f$ acts opposite to displacement throughout the path. The distance covered is $A_n + A_{n+1}$.
$$W_{fric} = -f (A_n + A_{n+1})$$
Using the Work-Energy Theorem ($\Delta E = W_{fric}$):
$$\frac{1}{2} k A_{n+1}^2 – \frac{1}{2} k A_n^2 = -f (A_n + A_{n+1})$$
Factor using $a^2 – b^2 = (a-b)(a+b)$:
$$\frac{1}{2} k (A_{n+1} – A_n)(A_{n+1} + A_n) = -f (A_n + A_{n+1})$$
Canceling $(A_{n+1} + A_n)$ from both sides:
$$\frac{1}{2} k (A_{n+1} – A_n) = -f$$
$$A_n – A_{n+1} = \frac{2f}{k}$$
Thus, the amplitude decreases by $\Delta A = \frac{2f}{k}$ in every half-cycle.
Calculating the decay value:
$$\Delta A = \frac{2 \times 2.5}{500} = \frac{5}{500} = 0.01 \text{ m} = 1.0 \text{ cm}$$
3. Sequence of Motion
The block starts at $x = +8.5$ cm. It oscillates back and forth, losing $1.0$ cm of amplitude per half-cycle.
- Start ($n=0$): $A_0 = 8.5$ cm.
- 1st Half Cycle: Travels to left. $A_1 = 8.5 – 1.0 = 7.5$ cm.
- 2nd Half Cycle: Travels to right. $A_2 = 7.5 – 1.0 = 6.5$ cm.
- 3rd Half Cycle: Travels to left. $A_3 = 6.5 – 1.0 = 5.5$ cm.
- 4th Half Cycle: Travels to right. $A_4 = 5.5 – 1.0 = 4.5$ cm.
- 8th Half Cycle: Travels to right. $A_4 = 1.5 – 1.0 = 0.5$ cm.
4. Analyzing the Stopping Condition
The block stops permanently when the restoring force of the spring at an extreme position is insufficient to overcome static friction ($k |x| \le f$).
$$|x| \le \frac{2.5}{500} = 0.5 \text{ cm}$$
Spring Relaxed: The spring becomes relaxed every time the block passes $x=0$. In 8 half-cycles, it crosses the center 8 times.
Number of times spring relaxed = 8