Two light springs of force constants k1 and k2 and a block of mass m are in one line AB on a smooth

Q: Why is the contact time with each spring $\frac{T}{2}$ and not some other fraction?

When the block hits the spring, it is at its maximum speed (equilibrium position). It compresses the spring to maximum compression (amplitude) and then returns to the equilibrium position where it loses contact. This represents exactly half of a complete SHM cycle (from $x=0$ to $x=A$ and back to $x=0$).

Q: Does the block's speed change while it is in the gap between the springs?

No, because the horizontal table is smooth (frictionless) and there are no horizontal forces acting on the block in the gap $CD$. Therefore, it moves with a constant velocity of $120\text{ cm/s}$.

Q: Why do we add the times instead of finding an equivalent spring constant?

An equivalent spring constant $k_{\text{eq}}$ is only defined when springs act simultaneously or continuously on the mass. Here, the springs act sequentially and are separated by a force-free region, so we must calculate the time duration of each phase independently and sum them up.

Q: What if the block was placed at some other initial position, would the time period change?

No! The time period of SHM is independent of the amplitude or initial position, and the time spent in the gap depends only on the constant speed, which is determined by the energy of the system. For a given maximum velocity, the time period remains constant.

Q: How do we get $2.82\text{ s}$ instead of $2.83\text{ s}$?

If you use the approximation $\sqrt{0.2/1.8} = \sqrt{1/9} \approx 0.33\text{ s}$ (rounding to two decimal places) and $\sqrt{0.2/3.2} = 0.25\text{ s}$, then the sum of the square roots is $0.58$. Multiplying this by $\pi \approx 3.14$ gives $1.82\text{ s}$. Adding the $1.0\text{ s}$ of free flight gives exactly $2.82\text{ s}$.

Question

Two light springs of force constants $k_1$ and $k_2$ and a block of mass $m$ are in one line $AB$ on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure.

The distance $CD$ between the free ends of the spring is $60	ext{ cm}$. If the block moves along $AB$ with a velocity $120	ext{ cm/s}$ in between the springs, calculate the period of oscillation of the block. (Take, $k_1 = 1.8	ext{ N/m}$, $k_2 = 3.2	ext{ N/m}$, $m = 200	ext{ g}$)

Sigma Percentile · Accepted Answer

### The Physics of Hybrid Oscillations

Imagine a block sliding back and forth on a frictionless horizontal table, bouncing between two springs.

This is not your standard simple harmonic motion (SHM) where a single spring is permanently attached to a mass.

Instead, it is a **hybrid oscillation**—a beautiful combination of constant-velocity free flight in the gap and half-cycles of SHM when the block is in contact with either spring.

To find the total time period of this non-standard oscillation, we must slice the motion into distinct, manageable phases.

### Slicing the Motion: The Four Phases

One complete cycle of oscillation consists of the block starting at one point, traveling to the other end, bouncing, and returning to its original state.

Let's trace this journey starting from the moment the block leaves the left spring at point $C$ moving to the right:

1. **Phase 1 (Free Flight to the Right):** The block travels from $C$ to $D$ across the gap of length $d = 60	ext{ cm}$ at a constant speed $v = 120	ext{ cm/s}$.

2. **Phase 2 (Bounce off Spring 2):** The block hits spring $k_2$ at point $D$, compresses it to maximum displacement, stops, and is pushed back out to $D$ with its velocity reversed.

3. **Phase 3 (Free Flight to the Left):** The block travels back from $D$ to $C$ across the gap at the same constant speed $v$.

4. **Phase 4 (Bounce off Spring 1):** The block hits spring $k_1$ at point $C$, compresses it, stops, and is pushed back out to $C$ with its velocity reversed, completing the cycle.

### Calculating the Free Flight Time

In the gap between the springs, there are no horizontal forces acting on the block because the table is perfectly smooth.

Therefore, the block moves with a **constant velocity**.

The time taken to cross the gap of distance $d = 0.6	ext{ m}$ at speed $v = 1.2	ext{ m/s}$ is:

$$t_{CD} = \frac{d}{v} = \frac{0.6	ext{ m}}{1.2	ext{ m/s}} = 0.5	ext{ s}$$

Since the return journey is identical, the total time spent in free flight during one complete cycle is:

$$t_{	ext{free}} = t_{CD} + t_{DC} = 0.5	ext{ s} + 0.5	ext{ s} = 1.0	ext{ s}$$

### The Spring Bounces: Half-Cycles of SHM

What happens when the block is in contact with a spring?

The block enters the spring at the equilibrium position (where the spring is unstretched) with maximum velocity, and it leaves the spring at the exact same position with its velocity reversed.

This represents exactly **half of a complete simple harmonic motion cycle**.

The contact time with a spring of constant $k$ is therefore half of the standard SHM time period:

$$t_{	ext{contact}} = \frac{T_{	ext{SHM}}}{2} = \frac{2\pi \sqrt{\frac{m}{k}}}{2} = \pi \sqrt{\frac{m}{k}}$$

Let's calculate this contact time for both springs:

* **For Spring 1 ($k_1 = 1.8	ext{ N/m}$):**

$$t_{	ext{contact}, 1} = \pi \sqrt{\frac{0.2}{1.8}} = \pi \sqrt{\frac{1}{9}} = \frac{\pi}{3}	ext{ s} \approx 1.047	ext{ s}$$

* **For Spring 2 ($k_2 = 3.2	ext{ N/m}$):**

$$t_{	ext{contact}, 2} = \pi \sqrt{\frac{0.2}{3.2}} = \pi \sqrt{\frac{1}{16}} = \frac{\pi}{4}	ext{ s} \approx 0.785	ext{ s}$$

### Putting It All Together

The total time period $T$ of the oscillation is the sum of the times spent in all four phases:

$$T = t_{	ext{free}} + t_{	ext{contact}, 1} + t_{	ext{contact}, 2}$$

$$T = 1.0 + \frac{\pi}{3} + \frac{\pi}{4} = 1.0 + \pi \left(\frac{7}{12}ight)$$

Using the approximation $\pi \approx 3.1416$:

$$T \approx 1.0 + 1.833 = 2.83	ext{ s}$$

Depending on the level of rounding used for the square roots (e.g., rounding $\sqrt{1/9} \approx 0.33$ and $\pi \approx 3.14$), the value is often written as **$2.82	ext{ s}$** in standard JEE answer keys.

This elegant problem shows how breaking a complex, non-linear motion into simple, symmetric parts makes it incredibly easy to solve!

Visualized Solution

Understanding the Setup

Slicing the Oscillation Cycle

Calculating Free Motion Time

Evaluating Free Motion Time

Understanding Spring Contact Time

Contact Time with Spring $k_{1}$

Evaluating Contact Time with Spring $k_{1}$

Contact Time with Spring $k_{2}$

Evaluating Contact Time with Spring $k_{2}$

Summing Up the Time Intervals

Calculating the Final Numerical Value

Exploring Variations

The Sigma Insight: Simple Harmonic Motion (SHM)

The Physics of Hybrid Oscillations

Slicing the Motion

Calculating the Free Flight Time

The Spring Bounces

Putting It All Together

Similar Questions

Two identical springs of spring constant $2 k$ are attached to a block of mass $m$ and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$ , then the frequency of oscillation of given body is