In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the

Q: Why does the frequency decrease when the mass is submerged in water?

When the mass is submerged in water, it experiences an upward buoyant force (upthrust). This upward force opposes gravity, reducing the net downward force on the wire. Since tension decreases, and frequency is directly proportional to the square root of tension, the fundamental frequency also decreases.

Q: Do we need to use the exact value of $g = 9.8\text{ m/s}^2$ or $10\text{ m/s}^2$?

No, because the acceleration due to gravity $g$ cancels out completely when we take the ratio of the new tension to the initial tension. This is a common feature in ratio-based physics problems.

Q: What if the mass was only partially submerged in water?

If the mass was only partially submerged, the buoyant force would be calculated using only the submerged volume ($V_{\text{sub}}$) instead of the total volume. The formula would be $F_B = V_{\text{sub}} \rho_w g$.

Q: Does the density of the wire or its length change when the mass is submerged?

No, the physical properties of the wire (length $l$ and linear mass density $\mu$) remain constant. Only the tension in the wire changes due to the buoyant force acting on the hanging mass.

Q: How did we get the density of water as $1000\text{ kg/m}^3$ when it wasn't given in the question?

The density of water is a standard physical constant equal to $1000\text{ kg/m}^3$ (or $1\text{ g/cm}^3$) in SI units. In JEE, such standard values are expected to be known by the student.

Question

In a sonometer wire, the tension is maintained by suspending a $50.7	ext{ kg}$ mass from the free end of the wire. The suspended mass has a volume of $0.0075	ext{ m}^3$. The fundamental frequency of vibration of the wire is $260	ext{ Hz}$. If the suspended mass is completely submerged in water, the fundamental frequency will become ...... $	ext{Hz}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a physics lab, watching a sonometer wire hum with a pure, steady tone.

This tone is the **fundamental frequency** of the wire, and it is governed by a delicate balance of physical properties: the length of the wire, its mass per unit length, and the tension stretching it tight.

In this classic problem, the tension is maintained by a hanging mass of $50.7	ext{ kg}$ suspended from the free end of the wire.

Initially, this mass hangs in the air, but we are going to submerge it completely in water.

How does this change the frequency? Let's dive into the physics.

### The Master Equation

The fundamental frequency $f$ of a stretched string of length $l$ and linear mass density $\mu$ is given by the formula:

$$f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$$

Since the length $l$ of the vibrating segment between the bridges and the linear density $\mu$ of the wire remain constant throughout the experiment, we can establish a direct proportionality:

$$f \propto \sqrt{T}$$

This means that the frequency of the sound produced is directly proportional to the square root of the tension in the wire.

Any change in tension will immediately reflect as a change in the frequency of the hum.

### The Intrusion of Archimedes

When the mass is hanging freely in the air, the tension $T$ in the wire is simply equal to the weight of the mass:

$$T = W = Mg = 50.7g$$

When the mass is completely submerged in water, it experiences an upward buoyant force (upthrust) $F_B$ according to Archimedes' Principle.

This buoyant force is equal to the weight of the water displaced by the submerged volume of the mass:

$$F_B = V ho_w g$$

Given that the volume of the mass $V = 0.0075	ext{ m}^3$ and the density of water $ho_w = 1000	ext{ kg/m}^3$, we can calculate the buoyant force:

$$F_B = 0.0075 	imes 1000 	imes g = 7.5g$$

This upward force reduces the net downward pull on the wire. The new tension $T'$ is the apparent weight of the mass:

$$T' = W - F_B = 50.7g - 7.5g = 43.2g$$

### The Mathematical Dance

Now, let's find the ratio of the new tension to the initial tension:

$$\frac{T'}{T} = \frac{43.2g}{50.7g} = \frac{43.2}{50.7}$$

To simplify this fraction, we multiply both the numerator and the denominator by $10$:

$$\frac{T'}{T} = \frac{432}{507}$$

Dividing both numbers by their common factor of $3$, we get:

$$\frac{T'}{T} = \frac{144}{169}$$

Notice how beautifully the numbers simplify! Both $144$ and $169$ are perfect squares ($12^2$ and $13^2$).

Now, we use our proportionality relation to find the new frequency $f'$:

$$\frac{f'}{f} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$

Given that the initial fundamental frequency $f = 260	ext{ Hz}$:

$$f' = 260 	imes \frac{12}{13} = 20 	imes 12 = 240	ext{ Hz}$$

Thus, the fundamental frequency of the wire when the mass is completely submerged in water becomes **$240	ext{ Hz}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Sonometer Setup

The Fundamental Frequency Formula

Establishing Proportionality

Submerging the Mass in Water

Calculating Apparent Tension

Substituting the Given Values

Simplifying the Tension Ratio

Simplifying the Fraction

Calculating the New Frequency

The Final Calculation

Exploring Further Variations

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Master Equation

The Intrusion of Archimedes

The Mathematical Dance

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