Question

The system shown in figure is in equilibrium. Pulley, springs and the strings are massless. The three blocks A, B and C have equal masses. x₁ and x₂ are extensions in the spring 1 and spring 2 respectively.

(a) Find the value of $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after spring I is cut.

(b) Find the value of $\left| \frac{d^2x_1}{dt^2} \right|$ and $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after string AB is cut.

(c) Find the value of $\left| \frac{d^2x_1}{dt^2} \right|$ and $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after spring 2 is cut.

Answer 1

a) $\left| \frac{d^2x_2}{dt^2} \right| = \frac{3g}{2}$, b) $\left| \frac{d^2x_1}{dt^2} \right| = 2g$, $\left| \frac{d^2x_2}{dt^2} \right| = 2g$, c) $\left| \frac{d^2x_1}{dt^2} \right| = 0$, $\left| \frac{d^2x_2}{dt^2} \right| = 2g

Answer 2

To solve this problem, we first need to determine the forces acting on each block and spring in the initial equilibrium state. Let the mass of each block (A, B, C) be $m$. Let $F_{s1}$ be the force in spring 1 and $F_{s2}$ be the force in spring 2. Let $T_{AB}$ be the tension in string AB, and $T$ be the tension in the string passing over the pulley.

1. Initial Equilibrium State Analysis:

• Block C: It is supported by spring 2.
  $F_{s2} - mg = 0 \implies F_{s2} = mg$.
  The extension in spring 2 is $x_2$, so $F_{s2} = k_2 x_2 = mg$.

• Block B: It is supported by string AB from above and pulls spring 2 from below.
  $T_{AB} - mg - F_{s2} = 0$.
  Substituting $F_{s2} = mg$: $T_{AB} - mg - mg = 0 \implies T_{AB} = 2mg$.

• Block A: It is supported by the pulley string from above and pulls string AB from below.
  $T - mg - T_{AB} = 0$.
  Substituting $T_{AB} = 2mg$: $T - mg - 2mg = 0 \implies T = 3mg$.

• Pulley & Spring 1: The string passes over the pulley. On the right side, it pulls block A with tension $T$. On the left side, it pulls spring 1. Since the pulley is massless, the tension in the string is uniform, $T$. Spring 1 is connected between the string and the ground. Thus, the force exerted by spring 1 is $F_{s1} = T$.
  $F_{s1} = 3mg$.
  The extension in spring 1 is $x_1$, so $F_{s1} = k_1 x_1 = 3mg$.

Summary of initial forces:

• $F_{s1} = 3mg$
• $F_{s2} = mg$
• $T_{AB} = 2mg$
• $T = 3mg$

Key Principles for "Immediately After Cutting":

• Forces in springs do not change instantaneously unless the spring itself is cut.
• Tension in strings changes instantaneously (becomes zero) if the string is cut.
• $d^2x/dt^2$ for a spring's extension means the relative acceleration of its two ends. If one end is fixed, it's the acceleration of the free end.

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(a) Find the value of $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after spring 1 is cut.

When spring 1 is cut:

• The force $F_{s1}$ becomes 0.
• The tension $T$ in the string over the pulley becomes 0.
• The force $F_{s2}$ in spring 2 remains $mg$.
• The string AB is still intact, so blocks A and B move together: $a_A = a_B$. Let's denote this common acceleration as $a_{AB}$.
• The acceleration of the extension of spring 2 is $d^2x_2/dt^2 = a_C - a_B$ (assuming $x_2$ increases when C moves down relative to B).

Let's apply Newton's second law (downwards positive):

• Block C: $mg - F_{s2} = m a_C$.
  Since $F_{s2} = mg$ (initial value): $mg - mg = m a_C \implies a_C = 0$.

• Block B: $mg + F_{s2} - T_{AB} = m a_B$.
  Since $F_{s2} = mg$: $mg + mg - T_{AB} = m a_B \implies 2mg - T_{AB} = m a_{AB}$. (Equation 1)

• Block A: $mg + T_{AB} - T = m a_A$.
  Since $T=0$: $mg + T_{AB} - 0 = m a_A \implies mg + T_{AB} = m a_{AB}$. (Equation 2)

Add Equation 1 and Equation 2:
$(2mg - T_{AB}) + (mg + T_{AB}) = m a_{AB} + m a_{AB}$
$3mg = 2m a_{AB}$
$a_{AB} = \frac{3g}{2}$.
So, $a_A = a_B = \frac{3g}{2}$ (downwards).

Now calculate $d^2x_2/dt^2$:
$\frac{d^2x_2}{dt^2} = a_C - a_B = 0 - \frac{3g}{2} = -\frac{3g}{2}$.
Therefore, $\left| \frac{d^2x_2}{dt^2} \right| = \frac{3g}{2}$.

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(b) Find the value of $\left| \frac{d^2x_1}{dt^2} \right|$ and $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after string AB is cut.

When string AB is cut:

• The tension $T_{AB}$ becomes 0.
• The forces $F_{s1}$ and $F_{s2}$ remain at their initial values: $F_{s1} = 3mg$ and $F_{s2} = mg$.
• The tension $T$ in the pulley string remains $3mg$.

Let's apply Newton's second law (downwards positive):

• Block C: $mg - F_{s2} = m a_C$.
  Since $F_{s2} = mg$: $mg - mg = m a_C \implies a_C = 0$.

• Block B: $mg + F_{s2} - T_{AB} = m a_B$.
  Since $F_{s2} = mg$ and $T_{AB} = 0$: $mg + mg - 0 = m a_B \implies 2mg = m a_B \implies a_B = 2g$ (downwards).

• Block A: $mg + T_{AB} - T = m a_A$.
  Since $T_{AB} = 0$ and $T = 3mg$: $mg + 0 - 3mg = m a_A \implies -2mg = m a_A \implies a_A = -2g$ (upwards).

Now calculate the accelerations of extensions:

• For spring 1: The bottom end of spring 1 is fixed. The top end is connected to the string. The string passes over the pulley and connects to block A. Since the string is inextensible and the pulley is massless, the acceleration of the top end of spring 1 is equal in magnitude and opposite in direction to the acceleration of block A. If $a_A$ is downwards, the spring end moves downwards. But $x_1$ is extension, so $x_1$ increases when the string pulls spring 1 downwards. The string is connected to the top end of spring 1. If A moves down by $y_A$, the string on the right moves down by $y_A$. The string on the left moves up by $y_A$. So the top end of spring 1 moves up by $y_A$.
  If $x_1$ is the extension, and the top end moves up, the extension decreases. So $d^2x_1/dt^2 = -a_{top\_end}$.
  The acceleration of the top end of spring 1 is $a_{top\_end} = -a_A$. (If $a_A$ is positive downwards, then $a_{top\_end}$ is positive upwards).
  So, $d^2x_1/dt^2 = -(a_A \text{ as calculated}) = -(-2g) = 2g$.
  Therefore, $\left| \frac{d^2x_1}{dt^2} \right| = 2g$.

• For spring 2: $\frac{d^2x_2}{dt^2} = a_C - a_B$.
  $\frac{d^2x_2}{dt^2} = 0 - 2g = -2g$.
  Therefore, $\left| \frac{d^2x_2}{dt^2} \right| = 2g$.

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(c) Find the value of $\left| \frac{d^2x_1}{dt^2} \right|$ and $\left| \frac{d^2x_2}{dt^2} \right|$ immediately after spring 2 is cut.

When spring 2 is cut:

• The force $F_{s2}$ becomes 0.
• The force $F_{s1}$ in spring 1 remains $3mg$.
• The tension $T_{AB}$ in string AB remains $2mg$.
• The tension $T$ in the pulley string remains $3mg$.

Let's apply Newton's second law (downwards positive):

• Block C: $mg - F_{s2} = m a_C$.
  Since $F_{s2} = 0$ (cut): $mg - 0 = m a_C \implies a_C = g$ (downwards).

• Block B: $mg + F_{s2} - T_{AB} = m a_B$.
  Since $F_{s2} = 0$ and $T_{AB} = 2mg$: $mg + 0 - 2mg = m a_B \implies -mg = m a_B \implies a_B = -g$ (upwards).

• Block A: $mg + T_{AB} - T = m a_A$.
  Since $T_{AB} = 2mg$ and $T = 3mg$: $mg + 2mg - 3mg = m a_A \implies 0 = m a_A \implies a_A = 0$.

Now calculate the accelerations of extensions:

• For spring 1: As established in part (b), $d^2x_1/dt^2 = -a_A$.
  Since $a_A = 0$: $\frac{d^2x_1}{dt^2} = -(0) = 0$.
  Therefore, $\left| \frac{d^2x_1}{dt^2} \right| = 0$.

• For spring 2: When spring 2 is cut, it no longer exerts an elastic force. The term "extension" for a cut spring is usually interpreted as the relative displacement between the points where it was attached. So, $d^2x_2/dt^2$ is the relative acceleration of C with respect to B.
  $\frac{d^2x_2}{dt^2} = a_C - a_B$.
  $\frac{d^2x_2}{dt^2} = g - (-g) = 2g$.
  Therefore, $\left| \frac{d^2x_2}{dt^2} \right| = 2g$.

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Summary of Results:

(a) Immediately after spring 1 is cut:
$\left| \frac{d^2x_2}{dt^2} \right| = \frac{3g}{2}$

(b) Immediately after string AB is cut:
$\left| \frac{d^2x_1}{dt^2} \right| = 2g$
$\left| \frac{d^2x_2}{dt^2} \right| = 2g$

(c) Immediately after spring 2 is cut:
$\left| \frac{d^2x_1}{dt^2} \right| = 0$
$\left| \frac{d^2x_2}{dt^2} \right| = 2g$