A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from the ceiling and

Question

A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has a force constant value of $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced along the length of the spring will be:

Accepted Answer

1. **Identify the Principle:** Since the only forces doing work are gravity and the spring force (both conservative), the total mechanical energy of the system is conserved [Class 11 Physics, Ch 6, Sec 5.8].
2. **Initial State:** The mass is released from rest ($v_i = 0$) from the unstretched position ($x = 0$). Let the reference level for gravitational potential energy be at this initial position. Thus, $K_i = 0$, $U_{g,i} = 0$, $U_{s,i} = 0$. Total Initial Energy $E_i = 0$.
3. **Final State (Maximum Extension):** The mass descends a distance $x_m$ (maximum extension). At this point, it momentarily comes to rest ($v_f = 0$), so kinetic energy $K_f = 0$. The spring is stretched by $x_m$, so Elastic Potential Energy $U_{s,f} = \frac{1}{2}kx_m^2$ [Class 11 Physics, Ch 6, Sec 5.9, Eq 5.15]. The mass is at a height $-x_m$ relative to the reference, so Gravitational Potential Energy $U_{g,f} = -Mgx_m$.
4. **Apply Conservation of Energy:**
   $$ E_i = E_f $$
   $$ 0 = \frac{1}{2}kx_m^2 - Mgx_m $$
   $$ Mgx_m = \frac{1}{2}kx_m^2 $$
   Since $x_m 
eq 0$, we can divide by $x_m$:
   $$ Mg = \frac{1}{2}kx_m $$
   $$ x_m = \frac{2Mg}{k} $$
   *(Note: This is different from the equilibrium position where $x_{eq} = Mg/k$. The mass oscillates around this equilibrium point)*.

Step-by-Step Solution

Practice All PHYSICS Questions