Question

In the figure shown a block of mass $m$ is attached in a light spring of spring constant $K$ and an identical spring hangs from ceiling. Initially lower spring is compressed in a state with compression equal to $\frac{3mg}{K}$ from natural length of spring. When block is released it strikes upper spring and sticks to it. Amplitude of oscillation (in cm) is given $mg = 10$ N and $K = 100\sqrt{7}$ N/m (roundoff answer to nearest integer)

$\frac{mg}{K}$ Equilibrium

Answer 1

5

Answer 2

The problem describes a block of mass $m$ attached to a lower spring of spring constant $K$ and an upper identical spring hanging from the ceiling. Initially, the lower spring is compressed by $\frac{3mg}{K}$ from its natural length. The block is released and moves upwards, strikes the upper spring, and sticks to it. We need to find the amplitude of oscillation after the block sticks to the upper spring.

Let's set the origin of the vertical coordinate $y$ at the natural length of the lower spring. Upward direction is positive. The initial position of the block is $y_i = -\frac{3mg}{K}$ (since the spring is compressed by $\frac{3mg}{K}$ from its natural length at $y=0$). The initial velocity is $v_i = 0$. The potential energy of the lower spring at position $y$ is $U_{lower}(y) = \frac{1}{2}K y^2$ (assuming natural length at $y=0$). The gravitational potential energy is $U_g(y) = mgy$. The initial total mechanical energy is $E_i = \frac{1}{2}mv_i^2 + U_{lower}(y_i) + U_g(y_i) = 0 + \frac{1}{2}K(-\frac{3mg}{K})^2 + mg(-\frac{3mg}{K}) = \frac{1}{2}K \frac{9m^2g^2}{K^2} - \frac{3m^2g^2}{K} = \frac{9m^2g^2}{2K} - \frac{6m^2g^2}{2K} = \frac{3m^2g^2}{2K}$.

The block strikes the upper spring. Let's assume the block strikes the upper spring when it reaches the natural length of the lower spring, i.e., at $y=0$. Let's find the velocity of the block at $y=0$ using conservation of energy. Let $v_0$ be the velocity at $y=0$. $E_{at y=0} = \frac{1}{2}mv_0^2 + U_{lower}(0) + U_g(0) = \frac{1}{2}mv_0^2 + 0 + 0 = \frac{1}{2}mv_0^2$. By conservation of energy, $E_{at y=0} = E_i$. $\frac{1}{2}mv_0^2 = \frac{3m^2g^2}{2K}$. $v_0^2 = \frac{3m^2g^2}{Km} = \frac{3mg^2}{K}$. Since the block moves upwards from a compressed state, $v_0 = \sqrt{\frac{3mg^2}{K}} = g\sqrt{\frac{3m}{K}}$.

When the block sticks to the upper spring at $y=0$, it is attached to both springs. The lower spring is attached to the ground at $y=0$ and the block. The upper spring is attached to the ceiling and the block. Let's assume the upper spring is at its natural length when its lower end is at $y=0$. So the upper spring is attached to the ceiling at $y = L_{0,upper}$, where $L_{0,upper}$ is the natural length of the upper spring. When the block is at position $y$, the force from the lower spring is $-Ky$. The force from the upper spring is due to its extension or compression from its natural length $L_{0,upper}$. The length of the upper spring when the block is at $y$ is $L_{upper} = |y_{ceiling} - y|$. If the upper spring is attached to the ceiling at $y_{ceiling}$, and its natural length is $L_{0,upper}$, and its lower end is at $y$, then the force is $K(y_{ceiling} - y - L_{0,upper})$ upwards. If the upper spring is at its natural length when its lower end is at $y=0$, then $y_{ceiling} - 0 = L_{0,upper}$, so $y_{ceiling} = L_{0,upper}$. The force from the upper spring is $K(L_{0,upper} - y - L_{0,upper}) = -Ky$ upwards. So the force is $-Ky$ downwards. The net force on the block at position $y$ is $F_{net} = F_{lower} + F_{upper} + F_g = -Ky - Ky - mg = -2Ky - mg$.

The equation of motion is $m\frac{d^2y}{dt^2} = -2Ky - mg$. Let's find the equilibrium position $y_{eq}$ where $F_{net} = 0$. $-2Ky_{eq} - mg = 0 \implies y_{eq} = -\frac{mg}{2K}$. Let $y' = y - y_{eq} = y + \frac{mg}{2K}$. Then $y = y' - \frac{mg}{2K}$. $m\frac{d^2(y' - \frac{mg}{2K})}{dt^2} = -2K(y' - \frac{mg}{2K}) - mg$. $m\frac{d^2y'}{dt^2} = -2Ky' + mg - mg = -2Ky'$. This is the equation for SHM with effective spring constant $K_{eff} = 2K$. The angular frequency is $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{2K}{m}}$.

The initial conditions for this SHM are at $t=0$, which is the moment the block sticks to the upper spring. At this moment, the position is $y=0$ and the velocity is $v_0$. The displacement from the equilibrium position at this moment is $y'(0) = y(0) - y_{eq} = 0 - (-\frac{mg}{2K}) = \frac{mg}{2K}$. The velocity at this moment is $v'(0) = v(0) = v_0 = g\sqrt{\frac{3m}{K}}$.

For SHM, the amplitude $A$ is related to the initial displacement and velocity by $A^2 = y'(0)^2 + (\frac{v'(0)}{\omega})^2$. $A^2 = (\frac{mg}{2K})^2 + (\frac{g\sqrt{\frac{3m}{K}}}{\sqrt{\frac{2K}{m}}})^2 = \frac{m^2g^2}{4K^2} + (\frac{g^2 \frac{3m}{K}}{\frac{2K}{m}}) = \frac{m^2g^2}{4K^2} + \frac{3m^2g^2}{2K^2} = \frac{m^2g^2}{4K^2} + \frac{6m^2g^2}{4K^2} = \frac{7m^2g^2}{4K^2}$. $A = \sqrt{\frac{7m^2g^2}{4K^2}} = \frac{\sqrt{7}mg}{2K}$.

We are given $mg = 10$ N and $K = 100\sqrt{7}$ N/m. $A = \frac{\sqrt{7} \times 10}{2 \times 100\sqrt{7}} = \frac{10\sqrt{7}}{200\sqrt{7}} = \frac{10}{200} = \frac{1}{20}$ m. We need the amplitude in cm. $A = \frac{1}{20}$ m $= \frac{1}{20} \times 100$ cm $= 5$ cm.

The amplitude of oscillation is 5 cm. We are asked to round off the answer to the nearest integer. Since 5 is an integer, the answer is 5.

Final check of assumptions:

1. The lower spring is attached to the ground.
2. The upper spring is attached to the ceiling.
3. The block is between the two springs.
4. The block strikes the upper spring at the natural length of the lower spring ($y=0$).
5. The upper spring is at its natural length when its lower end is at $y=0$. These assumptions are consistent with the figure and the problem statement.

The final answer is $\boxed{5}$.

Explanation of the solution:

1. Calculate the initial energy of the block when the lower spring is compressed by $\frac{3mg}{K}$ from its natural length.
2. Determine the velocity of the block when it reaches the position where it strikes the upper spring (assumed to be the natural length of the lower spring, $y=0$) using conservation of energy.
3. After sticking to the upper spring, the block is under the influence of both springs and gravity. Determine the effective spring constant of the system and the new equilibrium position.
4. Use the initial position and velocity at the moment of sticking (which is the start of the combined oscillation) relative to the new equilibrium position to calculate the amplitude of oscillation.
5. Convert the amplitude to cm and round off to the nearest integer.