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 $256Hz$. Keeping the stretching force constant and reducing the length of wire by $10cm$, the frequency becomes $320Hz$. Calculate the original length of wire.

Answer 1

Forced vibrations occur when the object is forced to vibrate at a particular frequency by a periodic input of force. If an object is being forced to vibrate at its natural frequency, resonance will occur and large amplitude vibrations can be observed. The resonant frequency is $f_0$.

Note that the frequency subscript matches the order of the overtone, not the order of the harmonic.

The fundamental frequency of vibration (the simplest mode with one vibrating segment) is given by the relation.

$f=\frac{1}{2L}\cdot \sqrt{\frac{T}{\mu }}$ where $f$ is the frequency in Hz, $T$ is the tension in the string in newton, $\mu$ is the mass per unit length of the string (linear density) in $kg/m$, and $L$ is the length of the string in meters.

Now, the linear density remains the same as there is no change in the material of the wire; also, the tension remains constant.

Hence, $256=\frac{1}{2L}\cdot \sqrt{\frac{T}{\mu }}$

or, $L=\frac{1}{2\left(256\right)}\cdot \sqrt{\frac{T}{\mu }}=\frac{1}{512}\cdot \sqrt{\frac{T}{\mu }}m$

Again, $320=\frac{1}{2(L-0\cdot 1)}\cdot \sqrt{\frac{T}{\mu }}$

or, $L-0.1=\frac{1}{2\left(320\right)}\cdot \sqrt{\frac{T}{\mu }}$

or, $\frac{1}{512}\cdot \sqrt{\frac{T}{\mu }}-0.1=\frac{1}{640}\cdot \sqrt{\frac{T}{\mu }}$

or, $\sqrt{\frac{T}{\mu }}[\frac{1}{512}-\frac{1}{640}]=0.1$

or, $\sqrt{\frac{T}{\mu }}\left[0.00039\right]=0.1$

or, $\sqrt{\frac{T}{\mu }}=256.41\sim 256$

Thus, $L=\frac{1}{512}\left(256\right)m$

$=0.5m=50cm$