## What is stretched string?

A stretched string is a wire which is made up of any metal that must have a larger length compared to its thickness. The ends of the wire are fixed, and when they oscillate or are plucked, it produces two reflected transverse waves of the same frequency and same amplitude. The waves or pulses are traveling in alternate directions and superimpose with one another. This waveform is named standing or stationary waves.

## Formation of stationary waves

When a wire is oscillated or plucked, it vibrated front and back and produces two transverse waves. An example is a guitar where its string ends are fixed. When the strings of the guitar is plucked, it forms two transverse waves which are traveling in alternate directions have the same wavelength λ, amplitude A, and frequency ν.

The equation of wave for the two transverse waves is written as,

${y}_{1}=A\mathrm{sin}(kx-\omega t)\phantom{\rule{0ex}{0ex}}{y}_{2}=-A\mathrm{sin}(kx+\omega t)$

Here,ω=2πν and k=(2π/λ)

The total resultant wave is,

$y={y}_{1}+{y}_{2}\phantom{\rule{0ex}{0ex}}y=A\mathrm{sin}(kx-\omega t)-(-A\mathrm{sin}(kx-\omega t\left)\right)\phantom{\rule{0ex}{0ex}}y=\left(2A\mathrm{sin}kx\right)\mathrm{cos}\omega t$

This equation is called a stationary wave equation.

Where,(2Asinkx) is the wave amplitude.

The amplitude depends upon the factor kx.

If $x=0,\frac{\lambda}{2},\frac{2\lambda}{2},\frac{3\lambda}{2}$....then, amplitude A= zero. This position is referred to as nodes.

If x=$0,\frac{\lambda}{4},\frac{3\lambda}{4},\frac{5\lambda}{4}.$....then, amplitude A=maximum. This position is referred to as antinodes.

The distance between a node and antinode is λ/4 and between consecutive two nodes and antinodes is λ/2.

## Harmonics and Overtone

A stretched string vibrates in different wavelengths and frequencies and creates stationary waves. This form of vibration is called harmonics.

If the string vibrates in one segment, it is known as a fundamental harmonic frequency or first harmonic. The increasing order of harmonics is known as overtones. It is the first integral of the fundamental frequency

If the string vibrates in two segments, then it is called second-harmonic or first overtone.

Consider the first wave from the image. It is first harmonic, the string is vibrating in a fundamental mode. The nodes are at the end and in the center anti node is present. A fully completed loop is formed.

Let the distance of separation of two successive nodes be λ_{2} and the string length is l.

$l=\frac{\lambda}{2}\phantom{\rule{0ex}{0ex}}\lambda =2l$

We know the velocity of wave v is,

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

Here, T = Tension and μ= linear density of the wave.

Therefore, the relationship between wavelength (λ) and the frequency (n) of the wave is,

$v=n\lambda \phantom{\rule{0ex}{0ex}}n=\frac{v}{\lambda}\phantom{\rule{0ex}{0ex}}n=\frac{1}{2l}\sqrt{\frac{T}{\mu}}$

This is the fundamental frequency of the wave string.

### Second harmonic or first overtone

Consider the second wave graph in the image. Here a node lies between two nodes and an antinode is placed at one-fourth of its distance from both ends. This form of wave is called second harmonic or first overtone.

Let the displacement between nodes be λ_{1},

$l={\lambda}_{1}\phantom{\rule{0ex}{0ex}}v={n}_{1}{\lambda}_{1}\phantom{\rule{0ex}{0ex}}{n}_{1}=\frac{v}{{\lambda}_{1}}\phantom{\rule{0ex}{0ex}}{n}_{1}=\frac{1}{l}\sqrt{\frac{T}{\mu}}$

Multiply and divide by 2 on the right side,

${n}_{1}=2\left[\frac{1}{2l}\sqrt{\frac{T}{\mu}}\right]\phantom{\rule{0ex}{0ex}}{n}_{1}=2n$

Similarly for third harmonic frequency or second overtone,

${n}_{2}=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\phantom{\rule{0ex}{0ex}}{n}_{2}=3{n}_{1}$

## Laws of stretched string along transverse wave

### First law: The law of length

_{ }The fundamental frequency (f) is inversely proportional to the square root of the resonating length of the string (l).

$f\alpha \frac{1}{\sqrt{l}}$ Where T and μ are constants.

### Second law: The law of tension

The fundamental frequency (f) is directly proportional to the square root of the tension (T) of the string.

$f\alpha \sqrt{T}$ Where l and μ are constants.

### Third law: The law of mass

The fundamental frequency of the string (f) is inversely proportional to the square root of the linear mass density.

$f\alpha \sqrt{\frac{1}{\mu}}$ Where T and l are constants.

## Wave speed on a stretched string

Let us consider a taut string, which is a string that is strong and straight and has tension. When it is in equilibrium, the tension is constant. In that string, take a small portion, the mass element is$\u2206m=\mu \u2206x.$ When it is at rest, the tension on both sides stays the same and opposite.

If the string is disturbed, it produces a transverse wave and proceeds in a positive x-direction. Because of restoring force, the element proceeds in the perpendicular direction to the wave and it does not travel in the x-direction. The tension ${F}_{T}$ acts both in the + x and –x -direction.

Assume the string is inclined concerning the horizontal x-axis which is small. The force of a small component, which is parallel to the string, is the addition of restoring force, and the tension forms in the string. The x-components nullify each other. Therefore the net force is the addition of the y-component.

The y-components of the force is,

$\mathrm{tan}{\theta}_{1}=\frac{-{F}_{1}}{{F}_{T}}and\phantom{\rule{0ex}{0ex}}\mathrm{tan}{\theta}_{2}=\frac{{F}_{2}}{{F}_{T}}$

The tan θ is equal to the slope of the function at a point and it is equal to the partial differentiation of y for x.

$\frac{{F}_{1}}{{F}_{T}}=-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}and\phantom{\rule{0ex}{0ex}}\frac{{F}_{2}}{{F}_{T}}=-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}$

The net force of the element is,

${F}_{net}={F}_{1}+{F}_{2}\phantom{\rule{0ex}{0ex}}{F}_{net}={F}_{T}\left[{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}\right]\phantom{\rule{0ex}{0ex}}$

Applying the second law of Newton, the net force is equal to the product of mass and acceleration. The linear density and mass of small elements are,

${F}_{T}\left[{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}\right]=\u2206ma\phantom{\rule{0ex}{0ex}}{F}_{T}\left[{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}\right]=\mu \u2206\times \frac{{\partial}^{2}y}{\partial {t}^{2}}$

Dividing the equation by ${F}_{T}\u2206x$

$\frac{{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}}{\u2206x}=\frac{\mu}{{F}_{T}}\left(\frac{{\partial}^{2}y}{\partial {t}^{2}}\right)$

By taking the limit, as∆x→0

$\underset{x\to 0}{\mathrm{lim}}\frac{{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{2}}-{\left(\frac{\partial y}{\partial x}\right)}_{{x}_{1}}}{\u2206x}=\frac{\mu}{{F}_{T}}\left(\frac{{\partial}^{2}y}{\partial {t}^{2}}\right)\phantom{\rule{0ex}{0ex}}\frac{{\partial}^{2}y}{\partial {x}^{2}}=\frac{\mu}{{F}_{T}}\left(\frac{{\partial}^{2}y}{\partial {t}^{2}}\right)$

As we know the linear differential equation is,

$\frac{{\partial}^{2}y(x,y)}{\partial {x}^{2}}=\frac{1}{{v}^{2}}\left(\frac{{\partial}^{2}y(x,t)}{\partial {t}^{2}}\right)$

Comparing both the equations,we get

$\frac{1}{{v}^{2}}=\frac{\mu}{{F}_{T}}$

The speed of a wave always relies on the tension and the linear density of the string.

## String instruments

The musical instruments which create sound by vibrating the stretched strings are called string instruments. By plucking, striking, or bowing it produces sound. Instruments like Veena, guitar, banjo, and sitar are known examples of plucked string. It is plucked by using a thumb or plectrum.

Violin, cello, viola, and double bass are some examples of bowing. By rubbing the string with a bow which is made up of stick with ribbons of horse tail hairs stretched at its ends. Another method of sound production is by striking the strings. The piano is an example of this type.

## Formulas

The fundamental frequency of the wave is given by the formula,

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

The speed of the wave or pulse of a string under the tension is given by,

$v=\frac{{F}_{T}}{\mu}$

## Context and Applications

This is an important topic in wave motion for all the students in graduates and postgraduates particularly for bachelors and masters in science (physics).

## Practice Problems

**Question 1: **A string of 0.5 m length has a mass of 0.30 g. If the tension produced in the string is 60 N, find the frequency.

- 96.36 Hz
- 385.66 Hz
- 316.22 Hz
- 485.87Hz

**Given data:**

Tension T= 60 N

Mass m=0.30 g

Length l=0.5 m

**Solution:**

The formula of frequency is ,

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

The linear density$\mu =\frac{m}{l}=\frac{0.30\times {10}^{-3}kg}{0.5m}=0.60\times {10}^{-3}kg/m$

Hence,

$n=\frac{1}{2\times 0.5}\sqrt{\frac{60N}{0.60\times {10}^{-3}kg/m}}\phantom{\rule{0ex}{0ex}}n=316.22Hz$

The frequency of the wave is 316.22 Hz.

**Answer: **The correct option is (c).

**Question 2: **The fundamental frequency is directly proportional to ______.

a. Tension

b. The square root of tension

c. Mass density

d. The square root of the length

**Answer:** The correct answer is (b).

**Explanation:** The fundamental frequency is directly proportional to the square root of the tension of the string.

FαT where length and mass density are constants.

**Question 3:** The speed of the transverse wave on a wire of length 30m is 50m/s. The tension on the wire is 300N. Find the mass of the wire.

- 3.6 kg
- 9.6 kg
- 4.2 kg
- 5.7 kg

**Answer**: The correct option is (a).

**Given data:**

Speed v=50 m/s

Length of the wire l =30 m

Tension T=300 N

**Explanation:**

The formula of the wave speed is,

$v=\frac{T}{\mu}\phantom{\rule{0ex}{0ex}}v=\frac{T}{{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}$

Where μ is the linear mass density, $\mu =\frac{m}{l}.$ Therefore,

$50=\sqrt{\frac{300m}{30}}\phantom{\rule{0ex}{0ex}}{50}^{2}=\frac{300m}{30}\phantom{\rule{0ex}{0ex}}2500=\frac{300}{m/30}\phantom{\rule{0ex}{0ex}}m=\frac{300\times 30}{2500}\phantom{\rule{0ex}{0ex}}m=3.6kg$

The mass of the wire is 3.6 kg

**Question 4:** The wavelength of the wave that travels through a medium is 20 m. If the frequency of the wave is 5 Hz, find the speed of the wave.

- 10 Hz
- 100 Hz
- 5 Hz
- 4 Hz

**Answer:** The correct option is (b).

**Given data:**

Wavelength λ=20 m

Frequency, n=5 Hz

**Explanation:**

The speed of the wave is determined from the formula,

$v=n\lambda \phantom{\rule{0ex}{0ex}}v=\left(5\right)\left(20\right)\phantom{\rule{0ex}{0ex}}v=100Hz$

The speed of the wave is 100 Hz.

**Question 5:** A string of mass 10 g has a length of 5 m. The speed of the wave produced in the string is 60 m/s. Find the tension in the string.

- 87 N
- 63 N
- 7.2 N
- 9.6 N

**Answer:** The correct option is (c).

**Given data:**

Mass m =10 g

length l=5 m

speed v=60 m/s

**Explanation:**

From the wave speed formula, we can calculate the tension of the string,

$v=\sqrt{\frac{T}{\mu}}\phantom{\rule{0ex}{0ex}}{v}^{2}=\frac{T}{{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$l$}\right.}}\phantom{\rule{0ex}{0ex}}{60}^{2}=\frac{T\times 5}{10\times {10}^{-3}kg}\phantom{\rule{0ex}{0ex}}T=\frac{3600\times 10\times {10}^{-3}kg}{5}\phantom{\rule{0ex}{0ex}}T=7.2N$

The tension of the string is 7.2 N.

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