Two identical strings X and Z made of same material have tensions T_X and T_Z in them. If their fund

Q: Why do we assume length and mass density are constant?

The problem explicitly states that the strings are 'identical' and 'made of the same material'. This implies they have the exact same physical dimensions (length) and the same mass per unit length (density).

Q: What exactly does 'fundamental frequency' mean?

The fundamental frequency is the lowest possible frequency of vibration for a standing wave on a string. It corresponds to the string vibrating in a single loop, forming one antinode in the middle and nodes at the two fixed ends.

Q: Why did we square the frequency ratio?

According to the formula, frequency is proportional to the square root of tension ($f \propto \sqrt{T}$). To isolate the ratio of tensions without the square root, we must mathematically square the ratio of the frequencies.

Q: Would the ratio change if the strings were vibrating in their second harmonic instead of the fundamental mode?

No, the ratio would remain exactly the same. The frequency of the $n$-th harmonic is $f_n = n \cdot f_1$. Since both strings would be in the same harmonic ($n=2$), the $n$ would cancel out when you take the ratio $\frac{f_{X,n}}{f_{Z,n}}$.

Q: Can we use this formula for very thick ropes or cables?

This formula assumes an ideal, perfectly flexible string. For very thick or rigid ropes, the stiffness of the material introduces slight deviations from this ideal behavior. However, for standard JEE physics problems, we always assume ideal strings unless stiffness is explicitly mentioned.

Question

Two identical strings $X$ and $Z$ made of same material have tensions $T_X$ and $T_Z$ in them. If their fundamental frequencies are $450	ext{ Hz}$ and $300	ext{ Hz}$ respectively, then the ratio of $T_X / T_Z$ is

Options:

(A) 1.25

(B) 2.25

(C) 0.44

(D) 1.5

Sigma Percentile · Accepted Answer

### The Symphony of Strings: Unraveling Frequency and Tension

Imagine you are standing in front of two identical guitars, and you pluck a single string on each. Even though the strings look exactly the same, one produces a higher-pitched sound than the other. Why does this happen? The secret lies in the invisible force pulling them tight: **tension**.

In this problem, we are given two identical strings, $X$ and $Z$, made of the exact same material. String $X$ vibrates with a fundamental frequency of $f_X = 450	ext{ Hz}$, while string $Z$ vibrates at $f_Z = 300	ext{ Hz}$. Our mission is to find the ratio of the tensions in these two strings, $\frac{T_X}{T_Z}$.

### The Master Equation

To bridge the gap between the sound we hear (frequency) and the physical force applied (tension), we rely on the classic formula for the fundamental frequency of a taut string:

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

Here, $l$ represents the vibrating length of the string, $T$ is the tension, and $\mu$ is the linear mass density (mass per unit length).

Now, let's look at the clues hidden in the problem statement. We are told the strings are **identical** and made of the **same material**. This is a massive hint! It means that both the length $l$ and the linear mass density $\mu$ are exactly the same for both strings $X$ and $Z$.

Because $\frac{1}{2l}$ and $\mu$ are constants in this scenario, we can strip away the clutter and reveal the core mathematical relationship:

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

This tells us that the frequency is directly proportional to the square root of the tension. If you want a higher pitch, you need to crank up the tension!

### Setting Up the Ratio

Since we are looking for a ratio, let's use our proportionality to set up an equation comparing string $X$ to string $Z$:

$$\frac{f_X}{f_Z} = \sqrt{\frac{T_X}{T_Z}}$$

We want to isolate the tension ratio, $\frac{T_X}{T_Z}$. To do this, we must eliminate the square root by squaring both sides of the equation:

$$\frac{T_X}{T_Z} = \left(\frac{f_X}{f_Z}ight)^2$$

### Final Calculation

Now comes the satisfying part—plugging in the numbers. We know $f_X = 450	ext{ Hz}$ and $f_Z = 300	ext{ Hz}$. Let's substitute these into our rearranged equation:

$$\frac{T_X}{T_Z} = \left(\frac{450}{300}ight)^2$$

Before we square anything, let's be smart and simplify the fraction inside the parentheses. Both $450$ and $300$ are divisible by $150$:

$$\frac{450}{300} = \frac{3}{2}$$

Now, we simply square this reduced fraction:

$$\frac{T_X}{T_Z} = \left(\frac{3}{2}ight)^2 = \frac{9}{4}$$

Finally, converting the fraction $\frac{9}{4}$ into a decimal gives us our answer:

**$$\frac{T_X}{T_Z} = 2.25$$**

And there we have it! The tension in string $X$ is $2.25$ times greater than the tension in string $Z$. This beautifully illustrates how a relatively small change in frequency requires a much larger proportional change in tension.

Animated Solution for Physics - Waves: Two identical strings $X$ and $Z$ made of same material have tensions $T_{X}$ and $T_{Z}$ in them. If their fundamental frequencies are $450 Hz$ and $300 Hz$ respectively, then the ratio of $T_{X} / T_{Z}$ is

Select Answer:

Visualized Solution

Visualizing the Vibrating Strings

The Fundamental Frequency Formula

Identifying the Constants

Setting Up the Ratio

Substituting the Values

Final Calculation

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Setting Up the Ratio

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Vibrating Strings

The Fundamental Frequency Formula

Identifying the Constants

Setting Up the Ratio

Substituting the Values

Final Calculation

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Setting Up the Ratio

Final Calculation

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes A and B is

Tube $A$ has both ends open while tube $B$ has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes $A$ and $B$ is