NLM O17

Step 2: Case I – String Remains Taut ($T \ge 0$)

Assume accelerations are equal: $a_1 = a_2 = a$. Equations of motion:

1. For A (upward): $T + k_1 x – mg = ma$
2. For B (downward): $mg + k_2 x – T = ma$

Adding the equations to eliminate $T$: $$ (k_1 + k_2) x = 2ma \implies a = \frac{(k_1 + k_2)x}{2m} $$ Now, solve for Tension $T$ using equation (2): $$ T = m(g – a) + k_2 x = mg + k_2 x – \frac{k_1 + k_2}{2} x $$ $$ T = mg – \frac{k_1 – k_2}{2} x $$ The string remains taut if $T \ge 0$: $$ mg \ge \frac{k_1 – k_2}{2} x \implies k_1 – k_2 \le \frac{2mg}{x} $$ $$ k_1 \le k_2 + \frac{2mg}{x} $$ Under this condition, $a_1 = a_2$. This matches Option (d).

Step 3: Case II – String Goes Slack

If $k_1 > k_2 + \frac{2mg}{x}$, the calculation yields a negative tension, which is impossible. The string goes slack ($T=0$). The blocks move independently:

• Block A: Upward force $k_1 x – mg$.
  $a_1 = \frac{k_1 x – mg}{m} = \frac{k_1 x}{m} – g$
• Block B: Downward force $k_2 x + mg$.
  $a_2 = \frac{k_2 x + mg}{m} = \frac{k_2 x}{m} + g$

We check if $a_1 > a_2$: $$ \left( \frac{k_1 x}{m} – g \right) > \left( \frac{k_2 x}{m} + g \right) $$ $$ (k_1 – k_2) \frac{x}{m} > 2g $$ $$ k_1 – k_2 > \frac{2mg}{x} $$ This inequality is exactly the condition for the string going slack. Thus, if the string is slack, $a_1 > a_2$. This matches Option (c).