A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscill

Q: Why is the length of the rod taken as $2L$ instead of $L$?

The problem explicitly states the length of the rod is $2L$. It's crucial to read the variables carefully, as using $L$ for the total length would lead to an incorrect moment of inertia calculation.

Q: How did we get the moment of inertia of the rod as $ML^2/3$?

The standard formula for a rod of mass $M$ and length $x$ about its center is $Mx^2/12$. Substituting $x = 2L$, we get $M(2L)^2/12 = 4ML^2/12 = ML^2/3$.

Q: Why do we add the moment of inertia of the two masses directly to the rod's moment of inertia?

Moment of inertia is a scalar quantity (more precisely, a tensor, but for a fixed axis of rotation, it acts like a scalar). The total moment of inertia of a composite system about a given axis is simply the algebraic sum of the moments of inertia of its individual components about that same axis.

Q: What does a 20% reduction in frequency mean mathematically?

A 20% reduction means the new frequency is the old frequency minus 20% of the old frequency. So, $\nu_2 = \nu_1 - 0.20\nu_1 = 0.80\nu_1$.

Q: Does the torsional constant $k$ change when we add the masses?

No, the torsional constant $k$ is a property of the suspension wire (its material, length, and radius). Adding masses to the rod does not change the properties of the wire, so $k$ remains constant.

Question

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to

Options:

(A) 0.57

(B) 0.37

(C) 0.77

(D) 0.17

Sigma Percentile · Accepted Answer

### The Physics of a Torsional Pendulum

Imagine a uniform rod suspended exactly at its midpoint by a wire. When you twist it slightly and let go, it starts oscillating back and forth. This is a classic **torsional pendulum**. The frequency of a torsional pendulum depends on two fundamental properties: the stiffness of the wire, called the torsional constant $k$, and the rod's resistance to being twisted, which is its moment of inertia $I$.

The mathematical relationship governing this motion is given by the frequency formula:

$$
u = \frac{1}{2\pi}\sqrt{\frac{k}{I}}$$

Notice that the frequency is inversely proportional to the square root of the moment of inertia. If the system becomes "heavier" to twist (higher $I$), it will oscillate more slowly (lower $
u$).

### Analyzing the Initial Setup

Let's figure out the initial moment of inertia. We have a rod of mass $M$ and total length $2L$. The standard formula for the moment of inertia of a uniform rod about an axis passing through its center is $\frac{	ext{mass} 	imes 	ext{length}^2}{12}$.

Substituting our specific values, we get:

$$I_1 = \frac{M(2L)^2}{12} = \frac{M(4L^2)}{12} = \frac{ML^2}{3}$$

### The Effect of Adding Masses

Next, we alter the system by attaching two small masses, each of mass $m$, to the rod. They are placed exactly at a distance of $L/2$ from the center on both sides.

The new moment of inertia will be the rod's original inertia plus the inertia contributed by these two masses. Since each mass acts as a point particle at a distance $r = L/2$, its moment of inertia is $mr^2$. Because there are two identical masses, we multiply this by two:

$$I_2 = \frac{ML^2}{3} + 2 	imes m\left(\frac{L}{2}ight)^2$$

$$I_2 = \frac{ML^2}{3} + \frac{mL^2}{2}$$

### The Master Equation

The problem states a very crucial detail: adding these masses reduces the oscillation frequency by $20\%$. This means the new frequency is $80\%$ of the original frequency, or $
u_2 = 0.8
u_1$.

Since frequency is inversely proportional to the square root of the moment of inertia, we can set up a neat ratio:

$$\sqrt{\frac{I_1}{I_2}} = 0.8$$

Let's get rid of that square root by squaring both sides. $0.8^2$ is $0.64$, which is equivalent to the fraction $\frac{16}{25}$. Now, we substitute our expressions for $I_1$ and $I_2$:

$$\frac{\frac{ML^2}{3}}{\frac{ML^2}{3} + \frac{mL^2}{2}} = \frac{16}{25}$$

Notice how every term contains an $L^2$? We can completely cancel out $L^2$ from the numerator and the denominator, leaving us with a much cleaner equation involving only the masses:

$$\frac{\frac{M}{3}}{\frac{M}{3} + \frac{m}{2}} = \frac{16}{25}$$

### Final Calculation

Now it's just simple algebra. Let's cross-multiply to solve for the ratio:

$$25 	imes \frac{M}{3} = 16 	imes \left(\frac{M}{3} + \frac{m}{2}ight)$$

Expanding the right side gives:

$$\frac{25M}{3} = \frac{16M}{3} + 8m$$

Subtracting $\frac{16M}{3}$ from $\frac{25M}{3}$ leaves us with $\frac{9M}{3}$, which simplifies to $3M$:

$$3M = 8m$$

We are at the finish line! We need the ratio of the small mass $m$ to the large mass $M$. Rearranging our equation, we get:

$$\frac{m}{M} = \frac{3}{8} = 0.375$$

Looking at our options, this is closest to **$0.37$**. And that is our final answer!

Select Answer:

Visualized Solution

The Torsional Pendulum Setup

Frequency of Torsional Oscillation

Initial Moment of Inertia

Adding Masses

Frequency Reduction Condition

Substituting Moment of Inertia

Solving for Mass Ratio

Final Answer

The Sigma Insight: Simple Harmonic Motion (SHM)

The Physics of a Torsional Pendulum

Analyzing the Initial Setup

The Effect of Adding Masses

The Master Equation

Final Calculation

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