Standing Sound Waves

Standing sound waves are a fascinating phenomenon that occurs when two identical sound waves travelling in opposite directions interfere with each other, creating a pattern of nodes and antinodes. This creates the impression that the wave is "standing still" rather than moving through the medium. These waves are not just a theoretical concept but have practical applications in real life. For example, musical instruments like guitars and flutes rely on standing waves to produce harmonious sounds. When a guitar string is plucked, it vibrates, creating standing waves that determine the pitch of the note. Similarly, standing sound waves are crucial in acoustics, influencing the design of concert halls to enhance sound quality by controlling how sound waves interact within the space. In this article, we will understand how standing sound waves help us grasp how sound is produced, manipulated, and optimized in various everyday scenarios.

Standing Waves

When two sets of progressive waves of the same type (both longitudinal or both transverse) having the same amplitude and the same time period or frequency or wavelength travelling along the same straight line with the same speed in opposite directions superimpose, a new set of waves are formed. These are called stationary waves.

Some of the Characteristics of Standing Waves

(1) In this the disturbance is confined to a particular region between the starting point and the reflecting point of the wave.
(2) In this there is no forward motion of the disturbance from one particle to the adjoining particle and so on, beyond this particular region.
(3) The total energy in the stationary wave is twice the energy of each incident and reflected wave. But there is no flow or transference of energy along the stationary wave.
(4) Points in a standing wave, which are permanently at rest. These are called nodes. The distance between two consecutive nodes is $\frac{\lambda }{2}$

(5) The Points on the standing wave having maximum amplitude are known as antinodes. The distance between two consecutive antinodes is also $\frac{\lambda }{2}$
(6) All the particles execute simple harmonic motion about their mean position (except those at nodes) within the same time period.

Note - In standing waves, if the amplitude of component waves is not equal. The resultant amplitude at nodes will not be zero. It will be a minimum. Because of this, some energy will pass across nodes and waves will be partially standing.

Let us take an example to understand and derive the equation of standing wave

Let us take a string and when a string is under tension and set into vibration, transverse harmonic waves propagate along its length. If the length of the string is fixed, reflected waves will also exist. These incident and reflected waves will superimpose to produce transverse stationary waves in a string

Incident wave $y_1=a\sin \frac{2\pi }{\lambda }(vt+x)$
Reflected wave $y_2=a\sin \frac{2\pi }{\lambda }[(vt-x)+\pi ]=-a\sin \frac{2\pi }{\lambda }(vt-x)$

Now we can apply the principle of superposition to this and get

$y=y_1+y_2=2a\cos \frac{2\pi vt}{\lambda }\sin \frac{2\pi x}{\lambda }$

Standing Wave in a Closed Organ Pipe

Organ pipes are musical instruments which are used for producing musical sound by blowing air into the pipe. In this longitudinal stationary waves are formed due to superimposition of incident and reflected longitudinal waves.

A closed organ pipe is a cylindrical tube having an air column with one end closed. Sound waves enter from a source vibrating near the open end. An ongoing pressure wave gets reflected from the fixed end. This inverted wave is again reflected at the open end. After two reflections, it moves towards the fixed end and interferes with the new wave sent by the source in that direction. The twice-reflected wave has travelled an extra distance of $2l$ causing a phase advance of $\frac{2\pi }{\lambda }\cdot 2l=\frac{4\pi l}{\lambda }$
Similarly at open ends, the twice reflected wave suffered a phase change of $\pi$ at the open end.

So the phase difference is $\delta =\frac{4\pi l}{\lambda }+\pi$. Also, the waves interfere constructively if the phase difference is $2n\pi$

$\begin{array}{rl}& \frac{4\pi l}{\lambda }+\pi =2n\pi \\ & l=(2n-1)\frac{\lambda }{4}\end{array}$

Here n = 1,2,3... But if we take n = 0,1,2,.... then the above equation can also be written as $l=(2n-1)\frac{\lambda }{4}$

So, the frequency can be written as $\nu =\frac{v}{\lambda }=\frac{v\cdot (2n-1)}{4l}$

Equation of standing wave is given by and explained earlier $=y=2a\cos \frac{2\pi t}{\lambda }\sin \frac{2\pi x}{\lambda }$
As, general formula for wavelength defined earlier $=\lambda =\frac{4L}{(2n-1)}$
The minimum allowed frequency is obtained by putting $\mathrm{n}=1$

(1) First normal mode of vibration : $n_1=\frac{v}{4L}$

This is called fundamental frequency. The note so produced is called the fundamental note or first harmonic.
(2) Second normal mode of vibration : $n_2=\frac{v}{{\lambda }_2}=\frac{3v}{4L}=3n_1$

This is called the third harmonic or first overtone.
(3) Third normal mode of vibration : $n_3=\frac{5v}{4L}=5n_1$

This is called the fifth harmonic or second overtone.

Standing Waves in Open Organ Pipes

The general formula for wavelength $\lambda =\frac{2L}{n}$ where $n=1,2,3\dots \dots \dots$

Then the first normal mode of vibration is $n_1=\frac{v}{{\lambda }_1}=\frac{v}{2L}$

This is called fundamental frequency and the node so produced is called fundamental node or first harmonic.

(2) Second normal mode of vibration $n_2=\frac{v}{{\lambda }_2}=\frac{v}{L}=2\left(\frac{v}{2L}\right)=2n_1\Rightarrow n_2=2n_1$

This is called the second harmonic or first overtone.
(3) Third normal mode of vibration $n_3=\frac{v}{{\lambda }_3}=\frac{3v}{2L},n_3=3n_1$

This is called the third harmonic or second overtone.

Recommended Topic Video

Solved Examples Based on Standing Sound Waves

Example 1: The phase difference between the two particles situated on both sides of a node is

1) $0^{\circ }$
2) ${90}^{\circ }$
3) ${180}^{\circ }$
4) ${360}^{\circ }$

Solution:

Standing wave

Two identical waves travel in opposite directions in the same medium and combine to form a stationary wave.

All the particles vibrating in a given loop are in the same phase and particles in two consecutive loops are in opposite phase to each other.

So two particles situated on both the sides of node will be in opposite phases of vibration since they correspond to different but consecutive loops.

Hence, the answer is the option (3).

Example 2: $y=5\cos \frac{\pi x}{25}\sin 100\pi t$. A node will not occur at a distance x is equal to in a stationary wave ( answer in cms)

1) 25

2) 62.5

3) 12.5

4) 37.5

Solution:

The node in the standing wave

The positions at which the amplitude of oscillation has zero value are called nodes.

wherein

At nodes,

$$\begin{array}{rl}& \sin kx=0\\ & kx=0,\pi ,2\pi \dots \dots n\pi \end{array}$$
nodes will occur, where
$$\begin{array}{rl}& \frac{\pi X}{25}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\text{etc.}\\ & X=12.5,37.5,62.5\mathrm{etc}\end{array}$$

Hence, the answer is the option (1).

Example 3: Tube A has both ends open while tube B has one end closed, otherwise, they are identical. The ratio of the fundamental frequency of tubes A and B

1) $1:2$
2) $1:4$
3) $2:1$
4) $4:1$

Solution:

Closed organ pipe

A closed organ pipe is a cylindrical tube having an air column with one end closed.

wherein

Condition of constructive interference

$l=(2n+1)\frac{\lambda }{4}$

In tube $A,{\lambda }_A=2l$
In tube $\mathrm{B}{\lambda }_B=4l$

$\begin{array}{rl}& \therefore v_A=\frac{\nu }{{\lambda }_A}=\frac{\nu }{2l}\\ & \therefore v_B=\frac{\nu }{{\lambda }_B}=\frac{\nu }{4l}\\ & \frac{v_A}{v_B}=\frac{2}{1}=2:1\end{array}$

Hence, the answer is the option (3).

Example 4: A closed organ pipe has a length ‘l'. The air in it is vibrating in 3rd overtone with maximum amplitude 'a'. Find the amplitude at a distance of l /7 from the closed end of the pipe.

1) a

2) $\frac{a}{2}$

3) $\sqrt{3}\frac{a}{2}$

4) $\frac{a}{\sqrt{2}}$

Solution:

The frequency in a closed organ pipe

$\begin{array}{rl}& \nu =(2n+1)\frac{V}{4l}\\ & n=0,1,2,3\dots \dots \end{array}$
wherein

$\begin{array}{rl}& V=\text{velocity of sound wave}\\ & l=\text{length of pipe}\\ & n=\text{number of overtones}\end{array}$

The figure shows the variation of displacement of particles in a closed organ pipe for 3rd overtone.

For the third overtone $l=\frac{7\lambda }{4}$ or $\lambda =\frac{4l}{7}$ or $\frac{\lambda }{4}=\frac{l}{7}$

Hence the amplitude at P at a distance $\frac{l}{7}$ from the closed end is ‘a’ because there is an antinode at that point

Alternate

Because there is a node at x = 0 the displacement amplitude as a function of x can be written as

$A=a\sin kx=a\sin \frac{2\pi }{\lambda }X$

For the third overtone $l=\frac{7\lambda }{4}$ or $\lambda =\frac{4l}{7}$
$\therefore A=a\sin \frac{7\pi }{2l}\frac{l}{7}=a\sin \frac{\pi }{2}=a\text{at}x=\frac{l}{7}\Rightarrow A=a$

Hence, the answer is the option (1).

Summary

If two waves with the same frequency and amplitude are moving in opposite directions and interfere with each other, then standing waves will be formed. It causes an interference pattern to form, wherein some points stay fixed and others oscillate with maximum amplitude, called antinodes. Standing sound waves are very important for musical instruments because they determine the pitch and timbre of the sound produced. They are equally important in many engineering applications for the design of acoustic devices or in searching for an understanding of resonance in structures.