Import Question JSON

$\text{The motion of the block can be divided into two parts: a constant velocity motion and a Simple Harmonic Motion (SHM) motion.}$ \\ \text{Part 1: Constant velocity motion}$ \\ $\text{Initially, the block moves with a constant velocity } v_0 \text{ for a distance } 2l \text{ until it hits the spring with spring constant } k \text{. The time taken for this part is given by:}$ \\ $t_1 = \frac{\text{distance}}{\text{velocity}} = \frac{2l}{v_0}$ \\ \text{Part 2: Simple Harmonic Motion}$ \\ $\text{After the block hits the spring } k\text{, it compresses the spring until its velocity becomes zero. Then it returns back. The block will be under the influence of two springs, one with spring constant } k \text{ and the other with spring constant } 4k \text{.}$ \\ $\text{The equivalent spring constant for the two springs in parallel will be:} \\ k_{eq} = k + 4k = 5k$ \\ $\text{The time period of the SHM will be:} \\ T_{SHM} = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{m}{5k}}$ \\ $\text{The block moves from the point of contact to maximum compression and then returns to the point of contact. This motion is one half of the total SHM time period. So, the time taken for this part is:} \\ t_2 = \frac{1}{2} T_{SHM} = \frac{1}{2} (2\pi \sqrt{\frac{m}{5k}}) = \pi \sqrt{\frac{m}{5k}}$ \\ $\text{But the question states that the block is given velocity } v_0 \text{ to the right. The block moves } 2l \text{ distance, then it will compress the } k\text{ spring. After it returns to its initial position, it will not be under the influence of spring with spring constant } 4k\text{. So, the time period of the block will be calculated considering two parts. From the initial position, the block moves } l \text{ distance, then it will get an impulse from the right wall. The block starts SHM with spring constant } 4k\text{. The time taken for one full oscillation is } T = 2\pi \sqrt{\frac{m}{4k}} = \pi\sqrt{\frac{m}{k}}$ \\ $\text{The motion of the block is described by a constant velocity motion and a half-oscillation. The initial velocity } v_0 \text{ is given to the block. The block moves } 2l \text{ distance, then it will undergo SHM with spring constant } k\text{. Time taken for the first part is } t_1 = \frac{2l}{v_0}$. \\ $\text{The block then moves to the left and will undergo SHM with both springs. This is a complex problem. Let's reconsider the problem from another perspective.}$ \\ $\text{The motion of the block is a sequence of events. Initially, the block moves with constant velocity } v_0 \text{ for a distance } 2l \text{ to the right. The time taken for this is } \frac{2l}{v_0}$. \\ $\text{Once the block makes contact with the right spring (with spring constant } 4k\text{, as seen from the image on the right), it undergoes SHM with this spring. The time period of this SHM is } T = 2\pi \sqrt{\frac{m}{4k}} = \pi \sqrt{\frac{m}{k}}$. \\ $\text{The time taken for a half oscillation (from maximum compression to back to the equilibrium position) is } t_2 = \frac{1}{2} T = \frac{1}{2} \pi \sqrt{\frac{m}{k}}$. \\ $\text{Then the block travels a distance } 2l \text{ to the left with constant velocity } v_0 \text{. The time taken for this is } t_3 = \frac{2l}{v_0}$. \\ $\text{The block then comes into contact with the left spring (with spring constant } k\text{) and undergoes SHM with this spring. The time period of this SHM is } T' = 2\pi \sqrt{\frac{m}{k}}$. \\ $\text{The time taken for a half oscillation is } t_4 = \frac{1}{2} T' = \frac{1}{2} (2\pi \sqrt{\frac{m}{k}}) = \pi \sqrt{\frac{m}{k}}$. \\ $\text{The total time period of the motion is the sum of all these times:} \\ T_{total} = t_1 + t_2 + t_3 + t_4 \\ T_{total} = \frac{2l}{v_0} + \frac{1}{2} \pi \sqrt{\frac{m}{k}} + \frac{2l}{v_0} + \pi \sqrt{\frac{m}{k}} \\ T_{total} = \frac{4l}{v_0} + \frac{3}{2} \pi \sqrt{\frac{m}{k}}$ \\ $\text{The spring constants in the image are } k \text{ for the left spring and } 4k \text{ for the right spring. The block is initially pushed to the right. The motion is split into four parts:} \\ \text{1. Free motion to the right for distance } 2l\text{. Time } t_1 = \frac{2l}{v_0}$. \\ $\text{2. SHM with spring constant } 4k \text{. This is a half oscillation. Time } t_2 = \frac{1}{2} T_1 = \frac{1}{2} (2\pi \sqrt{\frac{m}{4k}}) = \frac{\pi}{2}\sqrt{\frac{m}{k}}$. \\ $\text{3. Free motion to the left for distance } 2l \text{. Time } t_3 = \frac{2l}{v_0}$. \\ $\text{4. SHM with spring constant } k \text{. This is a half oscillation. Time } t_4 = \frac{1}{2} T_2 = \frac{1}{2} (2\pi \sqrt{\frac{m}{k}}) = \pi\sqrt{\frac{m}{k}}$. \\ $\text{The total time period is the sum of the times of these four motions:} \\ T_{total} = t_1 + t_2 + t_3 + t_4 = \frac{2l}{v_0} + \frac{\pi}{2}\sqrt{\frac{m}{k}} + \frac{2l}{v_0} + \pi\sqrt{\frac{m}{k}} \\ T_{total} = \frac{4l}{v_0} + (\frac{1}{2}+1)\pi\sqrt{\frac{m}{k}} = \frac{4l}{v_0} + \frac{3\pi}{2}\sqrt{\frac{m}{k}}$