A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes betwe

Q: Why does the frequency of the wire remain the same when we change the hanging mass?

The wire is undergoing forced oscillations driven by the tuning fork. In resonance, the wire's vibration frequency must match the natural frequency of the tuning fork, which is constant. Changing the mass only alters the wave speed, forcing the wire to adjust its wavelength (and thus the number of loops) to maintain resonance at that same frequency.

Q: What is the relationship between the number of antinodes and the tension in the wire?

From the formula $f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, since $f$, $L$, and $\mu$ are constant, the product $n \sqrt{T}$ must be constant. This means the number of antinodes $n$ is inversely proportional to the square root of the tension $T$ ($n \propto \frac{1}{\sqrt{T}}$).

Q: Why are the ends of the wire at the bridges always nodes?

The bridges act as rigid, unyielding supports that clamp the wire in place. Because the wire is physically constrained and cannot vibrate vertically at these contact points, the displacement is always zero, making them displacement nodes.

Q: What would happen to the number of loops if we used a thicker wire under the same conditions?

A thicker wire has a larger linear mass density ($\mu$). Since wave speed is $v = \sqrt{T/\mu}$, a larger $\mu$ decreases the wave speed. To maintain the same frequency $f = \frac{n v}{2L}$, the number of loops $n$ would have to increase to compensate for the slower wave speed.

Q: Can a sonometer wire vibrate with a non-integer number of loops?

No. For stable standing waves to form, the boundary conditions must be satisfied, meaning there must be nodes at both clamped ends. This requires the length of the wire $L$ to be an exact integer multiple of half-wavelengths ($L = n \frac{\lambda}{2}$), restricting $n$ to positive integers ($1, 2, 3, \dots$).

Question

A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9	ext{ kg}$ is suspended from the wire. When this mass is replaced by mass $M$, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is

Options:

(A) $25	ext{ kg}$

(B) $5	ext{ kg}$

(C) $12.5	ext{ kg}$

(D) $1/25	ext{ kg}$

Sigma Percentile · Accepted Answer

### Introduction to Sonometer Resonance

Imagine a tightly stretched guitar string. When you pluck it, it vibrates and produces a beautiful, crisp sound.

But what determines the pitch, or frequency, of that sound?

In physics, we study this using a classic laboratory apparatus called a **sonometer**.

A sonometer allows us to explore the relationship between the tension in a wire, its length, and the frequency of the standing waves formed on it.

In this problem, we are presented with a fascinating scenario where a sonometer wire is driven by a tuning fork of a fixed frequency.

By changing the hanging mass (which changes the tension), we observe a change in the number of loops (antinodes) formed on the wire.

Let's dive deep into the physics of this phenomenon and solve for the unknown mass $M$.

---

### The Physics of Standing Waves on a String

When a wave travels along a stretched string and reaches a fixed boundary, it reflects back.

The superposition of the incident wave and the reflected wave creates a **standing wave**.

Because the wire is clamped at the bridges of the sonometer, these boundary points cannot move.

Therefore, the ends of the vibrating segment of the wire must always be **displacement nodes**.

Between these nodes, the wire vibrates in a series of loops.

The points of maximum displacement are called **antinodes**.

If the wire vibrates in $n$ loops, it is vibrating in its $n$-th harmonic.

The length of one loop is exactly half a wavelength ($\frac{\lambda}{2}$).

Therefore, for a wire of length $L$ vibrating with $n$ loops, we have:

$$L = n \frac{\lambda}{2} \implies \lambda = \frac{2L}{n}$$

Using the wave equation $v = f \lambda$, we can express the frequency $f$ as:

$$f = \frac{v}{\lambda} = \frac{n v}{2L}$$

---

### The Role of Tension and Wave Speed

The speed $v$ of a transverse wave on a stretched string depends on two physical properties: the tension $T$ in the string and its linear mass density $\mu$ (mass per unit length).

This relationship is given by the famous formula:

$$v = \sqrt{\frac{T}{\mu}}$$

Substituting this expression for $v$ into our frequency equation gives us the master formula for sonometer resonance:

$$f = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

This equation is a beautiful bridge connecting the geometry of the vibration ($n$ and $L$), the physical properties of the string ($\mu$), and the external force ($T$).

---

### Analyzing Case 1: The $9	ext{ kg}$ Mass

In the first experiment, a mass of $9	ext{ kg}$ is suspended from the wire.

This hanging mass creates a tension $T_1$ in the wire due to gravity:

$$T_1 = m_1 g = 9g$$

Under this tension, the wire vibrates with **five antinodes**, which means it forms $5$ loops ($n_1 = 5$).

Let's substitute these values into our master equation:

$$f = \frac{5}{2L} \sqrt{\frac{9g}{\mu}}$$

This is our first mathematical relationship.

---

### Analyzing Case 2: The Unknown Mass $M$

In the second experiment, the $9	ext{ kg}$ mass is replaced by an unknown mass $M$.

This changes the tension in the wire to:

$$T_2 = Mg$$

With this new tension, the wire vibrates with **three antinodes**, forming $3$ loops ($n_2 = 3$).

Substituting these into the master equation gives:

$$f = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$$

This is our second mathematical relationship.

---

### Equating and Solving for $M$

Here is the key conceptual breakthrough: **the tuning fork is identical in both cases**.

Since the tuning fork drives the vibration, the frequency $f$ of the standing wave remains constant.

Furthermore, the positions of the bridges have not changed, so the length $L$ is also constant.

Since the frequency $f$ is the same, we can equate our two expressions:

$$\frac{5}{2L} \sqrt{\frac{9g}{\mu}} = \frac{3}{2L} \sqrt{\frac{Mg}{\mu}}$$

Now, let's simplify this equation.

Notice that the factor $\frac{1}{2L}$ appears on both sides, so we can divide both sides by it.

Similarly, the term $\sqrt{\frac{g}{\mu}}$ is common to both sides and can be factored out.

This leaves us with a remarkably simple equation:

$$5 \sqrt{9} = 3 \sqrt{M}$$

Since $\sqrt{9} = 3$, we substitute this in:

$$5 	imes 3 = 3 \sqrt{M}$$

Dividing both sides by $3$:

$$5 = \sqrt{M}$$

To solve for $M$, we square both sides of the equation:

$$M = 5^2 = 25	ext{ kg}$$

---

### Physical Intuition: Why Did the Mass Increase?

Let's think about this result intuitively.

When we went from $5$ loops to $3$ loops, the number of loops decreased.

Fewer loops mean a longer wavelength for each loop.

Since the frequency $f$ is fixed by the tuning fork, a longer wavelength requires a faster wave speed ($v = f \lambda$).

To get a faster wave speed, we must increase the tension in the wire ($v \propto \sqrt{T}$).

Therefore, the hanging mass must be significantly larger than the original $9	ext{ kg}$.

Our calculated value of $25	ext{ kg}$ perfectly aligns with this physical intuition!

Thus, the correct option is **(a)**.

Select Answer:

Visualized Solution

Understanding the Sonometer Setup

The Fundamental Formula for Standing Waves

Analyzing Case 1 with $9 kg$ Mass

Analyzing Case 2 with Mass $M$

Equating the Frequencies

Cancelling Common Terms

Solving for the Unknown Mass $M$

Final Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction to Sonometer Resonance

The Physics of Standing Waves on a String

The Role of Tension and Wave Speed

Analyzing Case 1

Analyzing Case 2

Equating and Solving for $M$

Physical Intuition

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