A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit le

Q: Why do we use $n=1, 3, 5, 14$ for the harmonics?

The problem explicitly states that the strings are vibrating in their $1^{\text{st}}, 3^{\text{rd}}, 5^{\text{th}}$, and $14^{\text{th}}$ harmonics respectively. The harmonic number directly corresponds to the value of $n$ in the frequency formula.

Q: Why is the frequency of all strings equal to $f_0$?

The problem states 'such that all the strings have same frequency'. Since String 1 is vibrating at $f_0$, every other string must also be vibrating at $f_0$.

Q: What happens to the $L_0$ and $\mu$ terms when we equate the frequencies?

Because $L_0$ and $\mu$ appear on both sides of the equation (inside and outside the square root), they act as common scaling factors and completely cancel out when you square both sides, leaving a simple ratio of tensions.

Q: Is it necessary to square both sides?

Yes, squaring both sides is the most efficient way to eliminate the square root over the tension $T$ and mass density $\mu$, making the algebraic isolation of $T_n$ much simpler and less prone to error.

Q: Can this method be applied to pipes or other instruments?

Absolutely! The core principle of equating frequencies and solving for an unknown variable is universal in wave mechanics, whether you are dealing with strings, open pipes, or closed pipes. Just ensure you use the correct boundary conditions and formulas.

Question

A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length $\mu$, $2\mu$, $3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$. It is found that in string-1 ($\mu$) at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0$.

The length of the string 1, 2, 3 and 4 are kept fixed at $L_0, \frac{3L_0}{2}, \frac{5L_0}{4}$ and $\frac{7L_0}{4}$, respectively. Strings 1, 2, 3 and 4 are vibrated at their $1^{	ext{st}}, 3^{	ext{rd}}, 5^{	ext{th}}$ and $14^{	ext{th}}$ harmonics, respectively such that all the strings have same frequency. The correct match for the tension in the four strings in the units of $T_0$ will be.

List-I:

P. String-1 ($\mu$)

Q. String-2 ($2\mu$)

R. String-3 ($3\mu$)

S. String-4 ($4\mu$)

List-II:

1. 1

2. 1/2

3. $1/\sqrt{2}$

4. $1/\sqrt{3}$

5. 3/16

6. 1/16

Sigma Percentile · Accepted Answer

### The Symphony of Strings

Imagine you are a master luthier, crafting a bizarre but beautiful musical instrument. This instrument doesn't just have one type of string; it has four, each with a different thickness (mass density) and length. When you pluck them, they don't just vibrate in their fundamental modes. Some vibrate in three loops, some in five, and one incredibly long string vibrates in a mesmerizing fourteen loops!

Yet, despite all this chaos, when you listen closely, they all produce the exact same pitch. They are in perfect harmony. How is this physically possible? The secret lies in the tension. By carefully tuning the tension of each string, we can force them to sing the same note. Let's dive into the mathematics of this beautiful phenomenon.

### The Master Equation

The frequency of any stretched string is governed by a single, elegant equation:

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

Here, $n$ is the harmonic number (the number of loops), $L$ is the length of the string, $T$ is the tension, and $\mu$ is the mass per unit length. This equation is our compass. Since all four strings have the same frequency, we can set up a series of equations equating the frequency of each string to the fundamental frequency of the first string, $f_0$.

### Setting the Baseline: String 1

Let's look at the first string. It's our reference point. It has a length of $L_0$, a mass density of $\mu$, and it's vibrating in its fundamental mode ($n=1$).

$$f_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}$$

This expression for $f_0$ is the golden standard. Every other string must match this exact value.

### Harmonizing String 2

Now, let's move to the second string. It's longer ($L = \frac{3L_0}{2}$), heavier ($\mu = 2\mu$), and vibrating in its third harmonic ($n=3$). Let's plug these into our master equation:

$$f_2 = \frac{3}{2 \left(\frac{3L_0}{2}ight)} \sqrt{\frac{T_2}{2\mu}}$$

Notice how the $3$ in the numerator and the $3$ in the denominator cancel out perfectly! This leaves us with:

$$f_2 = \frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}}$$

Since $f_2$ must equal $f_0$, we equate the two expressions:

$$\frac{1}{L_0} \sqrt{\frac{T_2}{2\mu}} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}$$

To solve for $T_2$, we square both sides to eliminate the square roots. The $L_0$ and $\mu$ terms gracefully cancel out, leaving:

$$\frac{T_2}{2\mu} = \frac{T_0}{4\mu}$$

$$T_2 = \frac{T_0}{2}$$

So, the tension in the second string is exactly half of the first string. The ratio is $1/2$.

### Tuning String 3

The third string is even more complex. It has a length of $\frac{5L_0}{4}$, a mass density of $3\mu$, and vibrates in the fifth harmonic ($n=5$).

$$f_3 = \frac{5}{2 \left(\frac{5L_0}{4}ight)} \sqrt{\frac{T_3}{3\mu}}$$

Once again, the math is kind to us. The $5$s cancel out, and the $2$ in the denominator simplifies with the $4$, giving:

$$f_3 = \frac{2}{L_0} \sqrt{\frac{T_3}{3\mu}}$$

Equating this to $f_0$ and squaring both sides:

$$\frac{4}{L_0^2} \left(\frac{T_3}{3\mu}ight) = \frac{1}{4L_0^2} \left(\frac{T_0}{\mu}ight)$$

$$4 \left(\frac{T_3}{3}ight) = \frac{T_0}{4}$$

$$T_3 = \frac{3T_0}{16}$$

The ratio for the third string is $3/16$.

### The Grand Finale: String 4

Finally, we reach the fourth string. It's the longest ($\frac{7L_0}{4}$), the heaviest ($4\mu$), and vibrates in a stunning $14^{	ext{th}}$ harmonic ($n=14$).

$$f_4 = \frac{14}{2 \left(\frac{7L_0}{4}ight)} \sqrt{\frac{T_4}{4\mu}}$$

Let's simplify the fraction. $14$ divided by $7$ is $2$, and the $2$s cancel out, leaving a $4$ in the numerator:

$$f_4 = \frac{4}{L_0} \sqrt{\frac{T_4}{4\mu}}$$

Equating to $f_0$ and squaring:

$$\frac{16}{L_0^2} \left(\frac{T_4}{4\mu}ight) = \frac{1}{4L_0^2} \left(\frac{T_0}{\mu}ight)$$

$$16 \left(\frac{T_4}{4}ight) = \frac{T_0}{4}$$

$$4 T_4 = \frac{T_0}{4} \implies T_4 = \frac{T_0}{16}$$

The ratio for the fourth string is $1/16$.

### Conclusion

We have successfully tuned our bizarre instrument! By systematically applying the wave frequency formula and carefully managing our algebra, we found the exact tensions required to bring all four strings into perfect harmony. The final matches are:

**String 1 $ightarrow$ 1**

**String 2 $ightarrow$ 1/2**

**String 3 $ightarrow$ 3/16**

**String 4 $ightarrow$ 1/16**

List-I

List-II

Select Matching Pairs:

Visualized Solution

$f = \frac{n}{2 L} \frac{T}{μ}$

$f_{0} = \frac{1}{2 L _{0}} \frac{T _{0}}{μ}$

$f_{2} = f_{0}$

$T_{2} = \frac{T _{0}}{2}$

$f_{3} = f_{0}$

$T_{3} = \frac{3 T _{0}}{16}$

$f_{4} = f_{0}$

$T_{4} = \frac{T _{0}}{16}$

\text{Final Match}

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Setting the Baseline

Harmonizing String 2

Tuning String 3

The Grand Finale

Conclusion

Similar Questions

List-I

List-II

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

A string of length $1 m$ and mass $2 \times 1 0^{- 5} kg$ is under tension $T$ . when the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is _______ Newton.

A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} kgm^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are : $420 Hz$ and $490 Hz$ . Then, $L$ in metres is

List-I

List-II

Select Matching Pairs:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f=2Ln​μT​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f0​=2L0​1​μT0​​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f2​=f0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​=2T0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f3​=f0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T3​=163T0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f4​=f0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T4​=16T0​​

\text{Final Match}

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Setting the Baseline

Harmonizing String 2

Tuning String 3

The Grand Finale

Conclusion

Similar Questions

List-I

List-II

$f = \frac{n}{2 L} \frac{T}{μ}$

$f_{0} = \frac{1}{2 L _{0}} \frac{T _{0}}{μ}$

$f_{2} = f_{0}$

$T_{2} = \frac{T _{0}}{2}$

$f_{3} = f_{0}$

$T_{3} = \frac{3 T _{0}}{16}$

$f_{4} = f_{0}$

$T_{4} = \frac{T _{0}}{16}$