Question

What are forced vibrations and resonance? Show that only odd harmonics are present in an air column vibrating in a pipe closed at one end.
A stretched wire emits a fundamental note of frequency of 256Hz. Keeping the stretching force constant and reducing the length of wire by 10cm, the frequency becomes 320Hz. Calculate the original length of the wire.

Answer 1

Hint: The forced vibrations in an object or a system occurs, when the object or the system is forced to vibrate at a particular frequency. To achieve this, force is input into the system periodically. When the amplitude of the vibrations produced by the system is highest, that frequency is known as the resonant frequency of the system and the system vibrating at that frequency is known to be at resonance. To prove that only odd harmonics are present in an air column vibrating in a pipe closed at one end, we will make a diagram of the system and solve it.
The fundamental frequency of vibration is given by the relation: $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}$.

Step by step solution:
Let’s start by defining the forced vibrations and resonance. The forced vibrations in an object or a system occurs, when the object or the system is forced to vibrate at a particular frequency. To achieve this, force is input into the system periodically. During this case, when the amplitude of the vibrations produced by the system is highest, that frequency is known as the resonant frequency of the system and the system vibrating at that frequency is known to be at resonance.
Now, let’s consider an organ pipe setup, with one end of the organ pipe closed. The length of the organ pipe is (L). Let’s consider (v) to be the speed of sound in air.

We know that when the sound waves travel down an air column in a cylindrical pipe which is closed at one end, then these waves are reflected back from the closed end. This reflected wave has a ${{180}^{0}}$phase difference against the incident wave. The interference between the incident and reflected waves, causes stationary waves to be formed in the air column with a node at the closed end and an antinode at the open end.
The fundamental mode or first harmonic is given by diagram (a). The distance between nodes and the consecutive antinode is \[\left( \dfrac{\lambda }{4} \right)\], where $(\lambda )$ is the wavelength of the sound wave.
Hence, $\lambda =4L$and the corresponding frequency of the wave (n) is; $n=\dfrac{v}{\lambda }\Rightarrow n=\dfrac{v}{4L}\to (1)$
Let’s see the next higher mode of vibration known as the first overtone. This wave contains 2 nodes and 2 antinodes as shown in diagram (b). The wavelength of this wave is $\left( {{\lambda }_{1}} \right)$given by; $L=\dfrac{3{{\lambda }_{1}}}{4}\Rightarrow {{\lambda }_{1}}=\dfrac{4L}{3}$ and the corresponding frequency $({{n}_{1}})$ is: ${{n}_{1}}=\dfrac{v}{{{\lambda }_{1}}}\Rightarrow {{n}_{1}}=\dfrac{v}{\left( \dfrac{4L}{3} \right)}\Rightarrow {{n}_{1}}=3\dfrac{v}{4L}\Rightarrow {{n}_{1}}=3n\to (2)$. Therefore, the frequency of the first overtone is 3 times the natural frequency that is an odd harmonic.
Similarly, for second harmonic, the wavelength of this wave would be $\left( {{\lambda }_{2}} \right)$ given by; $L=\dfrac{5{{\lambda }_{2}}}{4}\Rightarrow {{\lambda }_{2}}=\dfrac{4L}{5}$ and the corresponding frequency $({{n}_{2}})$ is: ${{n}_{2}}=\dfrac{v}{{{\lambda }_{2}}}\Rightarrow {{n}_{2}}=\dfrac{v}{\left( \dfrac{4L}{5} \right)}\Rightarrow {{n}_{2}}=5\dfrac{v}{4L}\Rightarrow {{n}_{2}}=5n\to (3)$. Therefore, the frequency of the second overtone is 5 times the natural frequency that is also an odd harmonic. Hence, only odd harmonics are present in an air column vibrating in a pipe closed at one end.
The fundamental frequency of vibration is given by the relation: $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}$, where (f) is the frequency, (T) is the tension of the string, (L) is the original length of the string and $\left( \mu \right)$ is the linear density of the wire. As per the problem, the tension of the wire is constant and since the wire is the same, hence the linear density of the wire also remains constant.
Therefore, $256=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}\Rightarrow L=\dfrac{1}{256(2)}\sqrt{\dfrac{T}{\mu }}\Rightarrow L=\dfrac{1}{512}\sqrt{\dfrac{T}{\mu }}\to (1)$.
Now, $320=\dfrac{1}{2(L-0.1)}\sqrt{\dfrac{T}{\mu }}\Rightarrow L-0.1=\dfrac{1}{320(2)}\sqrt{\dfrac{T}{\mu }}\Rightarrow L-0.1=\dfrac{1}{640}\sqrt{\dfrac{T}{\mu }}$.
Substituting in the value of (L) from the previous equation (1), we get; $\dfrac{1}{512}\sqrt{\dfrac{T}{\mu }}-0.1=\dfrac{1}{640}\sqrt{\dfrac{T}{\mu }}\Rightarrow 0.1=\sqrt{\dfrac{T}{\mu }}\left[ \dfrac{1}{512}-\dfrac{1}{640} \right]\Rightarrow 0.1=(0.00039)\sqrt{\dfrac{T}{\mu }}$.
That is; $\sqrt{\dfrac{T}{\mu }}=256.41\approx 256$. Putting in this value into equation (1), we get; $L=\dfrac{1}{512}\sqrt{\dfrac{T}{\mu }}\Rightarrow L=\dfrac{1}{512}\times 256\Rightarrow L=2m.$Hence, the original length of the wire is 2m.
Note: Every material has a resonant frequency at which it will produce highest amplitude oscillations. To understand, an example of resonance, you might have heard that people are never asked to march on a bridge because, a large number of people marching on a bridge may cause the bridge to resonate at its resonant frequency, which will cause all the particles of the bridge to oscillate at the highest amplitude damaging the bridge. However, when people are walking normally, it is more likely that people’s normal walking frequency don’t match with each other, but in the case of marching, all the people march at the same frequency.