In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of

Q: Why do we use the concept of reduced mass here?

Because both masses are free to move, they oscillate relative to each other and their common center of mass. By shifting our perspective to the center of mass frame, the mathematics simplifies beautifully. The system behaves exactly like a single block of mass $\mu$ (reduced mass) attached to a rigid wall.

Q: How do I know if the springs are in series or parallel?

Springs are in series when they are connected end-to-end, meaning the same restoring force passes through both of them, but their individual extensions add up to give the total extension. If they were connected side-by-side between the same two objects, they would be in parallel.

Q: Do I need to worry about the initial extension of the springs?

No. The angular frequency of a simple harmonic oscillator depends strictly on the inertial property (mass) and the restoring property (spring constant). The initial extension only determines the amplitude of the oscillation, not its frequency.

Q: Why did we convert the masses from grams to kilograms?

To ensure our final answer is in standard SI units (radians per second), all input variables must be in SI units. The spring constant is given in N/m (which involves kg), so the mass must be in kg to maintain dimensional consistency.

Q: What would happen if the surface wasn't frictionless?

If there were kinetic friction, the system would undergo damped harmonic motion. The amplitude would gradually decrease over time, and the angular frequency would be slightly lower than the natural undamped frequency we just calculated.

Question

In the reported figure, two bodies A and B of masses $200	ext{ g}$ and $800	ext{ g}$ are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ......... $	ext{rad/s}$ when $k = 20	ext{ N/m}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a frictionless horizontal table with two blocks, A and B, connected by a pair of springs. Block A has a mass of $200	ext{ g}$ and Block B has a mass of $800	ext{ g}$. The two springs connecting them have spring constants $k$ and $4k$, respectively, where $k = 20	ext{ N/m}$.

This is a classic **two-body oscillator**. Unlike a standard block-spring system where one end of the spring is attached to a rigid, immovable wall, here both ends are attached to movable masses. When displaced and released, both blocks will oscillate back and forth relative to their common center of mass.

To find the angular frequency $\omega$ of this system, analyzing the motion of each block separately using Newton's laws can be mathematically tedious. Instead, we can use a powerful physics trick: we can reduce this two-body problem into an equivalent **one-body problem**.

### The Master Equation: Reduced Mass

In the center of mass frame, a two-body system oscillates exactly like a single mass attached to a fixed wall. This single equivalent mass is called the **reduced mass**, denoted by $\mu$ or $m_{eq}$. The formula for reduced mass is the product of the two masses divided by their sum:

$$m_{eq} = \frac{m_A m_B}{m_A + m_B}$$

Before we plug in the numbers, we must ensure our units are strictly in the SI system. We convert the masses from grams to kilograms:
$m_A = 200	ext{ g} = 0.2	ext{ kg}$
$m_B = 800	ext{ g} = 0.8	ext{ kg}$

Now, substituting these values into our reduced mass formula:

$$m_{eq} = \frac{0.2 	imes 0.8}{0.2 + 0.8} = \frac{0.16}{1.0} = 0.16	ext{ kg}$$

### Equivalent Spring Constant

Next, we need to find the equivalent spring constant, $k_{eq}$, for the two springs. Notice how the springs are connected end-to-end. This means they are in **series**. When springs are in series, they share the same restoring force, but their extensions add up. The formula for the equivalent spring constant of two springs in series is similar to that of parallel resistors:

$$k_{eq} = \frac{k_1 k_2}{k_1 + k_2}$$

Substituting $k_1 = k$ and $k_2 = 4k$:

$$k_{eq} = \frac{k 	imes 4k}{k + 4k} = \frac{4k^2}{5k} = \frac{4k}{5}$$

We are given that $k = 20	ext{ N/m}$. Let's substitute this value to find the numerical value of $k_{eq}$:

$$k_{eq} = \frac{4 	imes 20}{5} = 16	ext{ N/m}$$

### Final Calculation

We have now completely transformed our complex system into a simple one: a single mass $m_{eq} = 0.16	ext{ kg}$ attached to a single spring $k_{eq} = 16	ext{ N/m}$. The angular frequency $\omega$ for a simple harmonic oscillator is given by:

$$\omega = \sqrt{\frac{k_{eq}}{m_{eq}}}$$

Let's substitute our calculated values into this final equation:

$$\omega = \sqrt{\frac{16}{0.16}}$$

$$\omega = \sqrt{100}$$

$$\omega = 10	ext{ rad/s}$$

**The angular frequency of the system is $10	ext{ rad/s}$.** What an elegant way to bypass complex differential equations!

Enter Numerical Value:

Visualized Solution

$m_{A}, m_{B}, k_{1}, k_{2}$

$m_{e q} = \frac{m _{A} m _{B}}{m _{A} + m _{B}}$

$m_{e q} = \frac{0.2 \times 0.8}{0.2 + 0.8}$

$m_{e q} = 0.16 kg$

$k_{e q} = \frac{k _{1} k _{2}}{k _{1} + k _{2}}$

$k_{e q} = \frac{4 k}{5}$

$k_{e q} = 16 N/m$

$ω = \frac{k _{e q}}{m _{e q}}$

$ω = 10 rad/s$

$What if?$

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Master Equation

Equivalent Spring Constant

Final Calculation

Similar Questions

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

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

Two particles A and B of equal masses are suspended from two massless springs of spring constants $k_{1}$ and $k_{2}$ , respectively. If the maximum velocities, during oscillations are equal, the ratio of amplitudes of A and B is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mA​,mB​,k1​,k2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } meq​=mA​+mB​mA​mB​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } meq​=0.2+0.80.2×0.8​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } meq​=0.16 kg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } keq​=k1​+k2​k1​k2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } keq​=54k​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } keq​=16 N/m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=meq​keq​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω=10 rad/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if?