Waves In Pipe:
a) Closed Pipe
An organ pipe which is closed at one end is called closed organ pipe. When air is blown into the pipe through the open end, the disturbance travelled towards the closed end and reflects back. As a result of which, stationary wave is produced in it due to superposition of waves. The air in contact with the closed end is at rest. So, there is always node. But, the open end air can vibrate with maximum amplitude. Hence, there is always antinode.
Modes of vibration:
Consider, a closed pipe of length βlβ. Let, βvβ be the velocity of sound in air.
a) 1st Mode of vibration:
In this mode of vibration, there is only one node and one antinode. If [latex]\lambda_1[/latex] be the wavelength of the wave, then,
[latex]l = \frac{\lambda_1}{4}[/latex]
Or, [latex]\lambda_1 = 4l[/latex]
We know, frequency ([latex]f_1[/latex]) = [latex]\frac{v}{\lambda_1}[/latex]
β΄ [latex]f_1 = \frac{v}{4l}[/latex] ………………………… (i)
This is called fundamental frequency of a closed pipe of length βlβ and is the minimum frequency, that can be obtained from given closed pipe.
b) 2nd Mode of vibration:
In this mode of vibration, there are 2 nodes and 2 antinodes. If [latex]\lambda_2[/latex] is the wavelength of the wave, then,
[latex]l = \frac{3\lambda_2}{4}[/latex]
Or, [latex]\lambda_2 = \frac{4l}{3}[/latex]
We know, frequency [latex]f_2 = \frac{v}{\lambda_2}[/latex]
β΄ [latex]f_2 = \frac{3v}{4l} = 3f_1[/latex] ……………..(ii) [β΅ from (i)]
where, [latex]f_1 = \frac{v}{4l}[/latex]
This is called third harmonic or first overtone.
c) 3rd Mode of vibration:
In this mode of vibration, there are 3 nodes and 3 antinodes. If [latex]\lambda_3[/latex] is the wavelength of the wave, then,
[latex]l = \frac{5\lambda_3}{4}[/latex]
Or, [latex]\lambda_3 = \frac{4l}{5}[/latex]
We know, frequency [latex]f_3 = \frac{v}{\lambda_3}[/latex]
β΄ [latex]f_3 = \frac{5v}{4l} = 5f_1[/latex] ……………………… (iii) [β΅ from (i)]
where, [latex]f_1 = \frac{v}{4l}[/latex]
This is called fifth harmonic or second overtone.
In general, the harmonics from a closed pipe are,
[latex]f_n = (2n-1)f_1, \quad \text{where, } n = 1, 2, 3,\dots[/latex]
Where, [latex]f_1 = \frac{v}{4l}[/latex]
From the relations above, we can conclude that the harmonics provided in a closed pipe are always an odd integral multiple of fundamental frequency.
Open Organ Pipe
An organ pipe which is open at both the ends is called open organ pipe. When air is blown into the pipe through one end, vibrations travel towards the other end. After deflection and superposition with the original wave, a stationary wave is formed inside the pipe. The air vibrates with maximum amplitude at both the end. So, the antinodes are formed there.
Mode of vibration:
Consider, an organ pipe of length βlβ and let, βvβ be the velocity of sound in air.
a) 1st Mode of vibration:
In this mode of vibration,
[latex]l = \frac{\lambda_1}{2}[/latex]
β΄ [latex]\lambda_1 = 2l[/latex]
So, [latex]f_1 = \frac{v}{\lambda_1} = \frac{v}{2l}[/latex] ……….. (i)
This is called fundamental frequency of an open pipe of length βlβ and is the minimum frequency that can be obtained from the given pipe.
b) 2nd Mode of vibration:
In this mode of vibration,
[latex]l = \lambda_2[/latex]
β΄ [latex]\lambda_2 = l[/latex]
So, [latex]f_2 = \frac{v}{\lambda_2} = \frac{v}{l} = 2.\frac{v}{2l} = 2f_1[/latex] ……………. (ii)
This is called second harmonics or first overtone of an open pipe.
c) 3rd Mode of vibration:
In this mode of vibration,
[latex]l = \frac{3\lambda_3}{2}[/latex]
β΄ [latex]\lambda_3 = \frac{2l}{3}[/latex]
So, [latex]f_3 = \frac{v}{\lambda_3} = \frac{3v}{2l} = 3.\frac{v}{2l} = 3f_1[/latex] ………………………. (iii)
This is called third harmonics or second overtone of an open pipe.
In general, the harmonics obtained from an open organ pipe are:
[latex]f_n = nf_1, \quad n = 1, 2, 3, \dots[/latex]
where, [latex]f_1 = \frac{v}{2l}[/latex]
Hence, in open organ pipe, all the odd and even harmonics can be obtained.
End Correction
In the derivation of harmonics of organ pipe, it is supposed that the antinode of stationary wave is exactly at the open end of the pipe. But, in practice, the antinode is formed a little away from the open end. The small extra distance between open end of the pipe and the point where the antinode is actually formed is called end correction (e or x). The effective length of the vibrating air column, therefore is slightly greater than actual length of the pipe. From theoretical consideration, Rayleigh determined it as [latex]0.3\Delta[/latex], where, [latex]\Delta[/latex] is the internal diameter of pipe.
In case of closed pipe,
[latex]l + e = \frac{\lambda}{4}[/latex]
Or, [latex]\lambda = 4(l+e)[/latex]
β΄ [latex]f = \frac{v}{\lambda} = \frac{v}{4(l+e)}[/latex]
In case of open pipe,
[latex]l + e + e = \frac{\lambda}{2}[/latex]
Or, [latex]\lambda = 2(l+2e)[/latex]
β΄ [latex]f = \frac{v}{\lambda} = \frac{v}{2(l+2e)}[/latex]
Waves in String
When one end of a stretched string is jerked up and down repeatedly, a transverse wave is produced which travels towards the fixed end. After reflection at the fixed end and their superposition results the transverse stationary wave within it.
Velocity of transverse wave in a stretched string (dimensional method):
The velocity of transverse wave along a stretched string depends on the tension applied (T), length of the string (l) and mass of the string βMβ, let,
[latex]v \propto T^x l^y M^z[/latex]
β΄ [latex]v = k T^x l^y M^z[/latex] ……………… (i)
Writing the dimensional formula on both sides, we get,
[latex][LT^{-1}] = [MLT^{-2}]^x [L]^y [M]^z[/latex]
[latex][LT^{-1}] = M^{x+z} L^{x+y} T^{-2x}[/latex]
Now, [latex]x + z = 0[/latex]
Or, [latex]x = -z[/latex]
And, [latex]x + y = 1[/latex]
Or, [latex]x = 1 β y[/latex]
β΄ [latex]-1 = -2x[/latex]
Or, [latex]x = \frac{1}{2}[/latex]
Then, [latex]y = \frac{1}{2}[/latex] and [latex]z = -\frac{1}{2}[/latex]
So, [latex]x = \frac{1}{2}, y = \frac{1}{2}\ and\ z = -\frac{1}{2}[/latex]
Now, [latex]v = k T^{1/2} l^{1/2} M^{-1/2}[/latex]
= k [latex]\sqrt{\frac{Tl}{M}}[/latex]
β΄ [latex]v = \sqrt{\frac{Tl}{M}} = \sqrt{\frac{T}{M/l}}[/latex]
β΄ [latex]v = \sqrt{\frac{T}{\mu}}[/latex]
where, [latex]\mu = M/l = \text{mass per unit length of wire}[/latex], its SI unit is [latex]\text{Kg m}^{-1}[/latex].
Velocity of Transverse Wave along a stretched string
If a wave travels through a medium, it must cause the particles of the medium to oscillate as it passes. There are two fundamentals that determine how fast a wave travels i.e. the mass and the elasticity of the medium. Let us consider, a stretched string as shown in figure above.
Let us take a single symmetrical pulse moving from left to right with speed, v. βdlβ be the length of the small segment of mass βdmβ of the pulse, that forms an arc of a circle of radius βRβ which subtends an angle 2ΞΈ at the centre O. Let us analyse the particle from the reference of frame of wave. This means βdlβ appears to move back with velocity v.
βTβ be the tension applied on the string, there are two components of the tension i.e. [latex]T\cos\theta[/latex] along horizontal and [latex]T\sin\theta[/latex] along vertical. The component [latex]T\cos\theta[/latex] cancels each other, the component [latex]T\sin\theta[/latex] acts towards the centre of the circle which is equal to centripetal force. So,
[latex]2T\sin\theta = F_c[/latex]
Or, [latex]2T\theta = F_c \quad (β΅ \text{If } \theta \text{ is small})[/latex] ………….. (i)
But, we have, centripetal force,
[latex]F_c = \frac{dm \cdot v^2}{R}[/latex]
Since, [latex]\mu[/latex] is the mass per unit length, [latex]\mu = \frac{dm}{dl}[/latex]
Or, [latex]dm = \mu dl[/latex]
So, [latex]F_c = \frac{\mu dlv^2}{R}[/latex] ……………….. (ii)
Also, arc length (dl) = [latex]R.2\theta [/latex]
So, eqn. (2) becomes,
[latex]F_c = \frac{\mu (R \cdot 2\theta)v^2}{R}[/latex]
Or, [latex]F_c = 2\mu v^2 \theta[/latex]
So, from eqns. (1) and (2), we get,
[latex]2T\theta = 2\mu v^2 \theta[/latex]
β΄ [latex]v = \sqrt{\frac{T}{\mu}}[/latex]
Which is required expression for the velocity of transverse wave along a stretched string.
Modes of vibration of a stretched string:
When a stretched string is jerked, transverse waves are produced, which travel towards the fixed end. They reflect towards the fixed end and superposition of them results a stationary wave in the string. Since, both the ends are fixed, there are always nodes.
Consider, a stretched string having length βlβ, mass per unit length β[latex]\mu[/latex]β and tension applied βTβ.
a) First Mode of vibration:
In this mode of vibration,
[latex]l = \frac{\lambda_1}{2}[/latex]
Or, [latex]\lambda_1 = 2l[/latex]
β΄ [latex]f_1 = \frac{v}{\lambda_1} = \frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex] ………………. (i)
This is called a fundamental frequency of stretched string having length βlβ, mass per unit length β[latex]\mu[/latex]β and tension βTβ applied.
b) Second Mode of vibration:
In this mode of vibration,
[latex]l = \lambda_2[/latex]
Or, [latex]\lambda_2 = l[/latex]
β΄ [latex]f_2 = \frac{v}{\lambda_2} = \frac{v}{l} = \frac{2}{2l}.v = \frac{2}{2l}\sqrt{\frac{T}{\mu}} = 2f_1[/latex] …………………… (ii)
This is called second harmonics or first overtone.
c) Third Mode of vibration:
In this mode of vibration,
[latex]l = \frac{3\lambda_3}{2}[/latex]
Or, [latex]\lambda_3 = \frac{2l}{3}[/latex]
β΄ [latex]f_3 = \frac{v}{\lambda_3} = \frac{3v}{2l} = \frac{3}{2l}\sqrt{\frac{T}{\mu}} = 3f_1[/latex] …………………… (iii)
This is called third harmonics or second overtone.
In general, [latex]f_n = nf_1, \quad \text{where, n = no. of harmonics}[/latex]
Laws of transverse vibration:
We have, [latex]f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}[/latex]
a) Law of length
We know, If T and [latex]\mu[/latex] are constant,
[latex]f \propto \frac{1}{l}[/latex], i.e., fundamental frequency in a stretched string is inversely proportional to the length of the string.
b) Law of tension
We have, if [latex]l[/latex] and [latex]\mu[/latex] are constant,
[latex]f \propto \sqrt{T}[/latex], i.e., fundamental frequency in a stretched string is directly proportional to the square root of its tension.
c) Law of mass
And, if T and [latex]l[/latex] are constant,
[latex]f \propto \frac{1}{\sqrt{\mu}}[/latex]
i.e., fundamental frequency in a stretched string is inversely proportional to the square root of its mass per unit length.
Experimental Verification of Laws of Transverse vibration of string:
Sonometer is used to verify the laws of transverse vibration of string experimentally. The experimental arrangement is shown in figure below.
To verify Law of Length
To verify this law, a tuning fork of known frequency is taken. The load of 1 kg is suspended in the string and resonating length is found out between the bridges C and D. Let, [latex]l_1[/latex] be the resonating length for this tuning fork. The process is repeated for tuning forks of different frequency. Suppose, [latex]l_2, l_3, l_4, l_5[/latex] are the resonating lengths of tuning forks of frequency [latex]f_2, f_3, f_4, f_5[/latex] respectively, such that,
[latex]f_1 \times l_1 = f_2 \times l_2 = f_3 \times l_3 = f_4 \times l_4 = f_5 \times l_5 [/latex] at constant T and [latex]\mu[/latex]
So,
[latex]fl = \text{Constant}[/latex]
Or, [latex]f \propto \frac{1}{l}[/latex]
A graph is plotted between f and 1/l, a straight line is obtained passing through origin as shown below. This verifies f β 1/l.
To verify law of Tension
To verify this law, the length between the bridges and mass per unit length are made constant. A known frequency tuning fork is stuck on rubber pad and used to vibrate the sonometer wire. The weight on the pan is changed until it vibrates with maximum amplitude. The tension is noted. This process is repeated for different tuning forks and corresponding tension is noted. A graph is plotted between f and [latex]\sqrt{T}[/latex], a straight line passing through origin is obtained. This verifies [latex]f \propto \sqrt{T}[/latex].
To verify Law of Mass
To verify this law, wires of same material but different diameters are taken along with a set of different frequencyβs tuning fork. A tuning fork of known frequency is stuck in rubber pad and is used to made one of the wires to vibrate with maximum amplitude keeping resonating length and tension constant. The frequency and mass per unit length [latex](\mu)[/latex] are noted. It is repeated changing the frequency and mass per unit length of the wire and a graph is plotted between f and [latex]\frac{1}{\sqrt{\mu}}[/latex], the graph passes through origin. This verifies [latex]f\propto \frac{1}{\sqrt{\mu}}[/latex].
