Question

A violin string 30.0cmlong with linear density 0.650g/mis placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880Hzand 1320Hzas the frequency of the oscillator is varied over the range 500-1500Hz. What is the tension in the string?

Short answer

The tension in the string is 45.3N

Step-by-step solution

The given data

1. Length of the string is $\left(L\right)=30.0cmor0.3m.$
2. Linear density,$\left(\mu \right)=0.650g/m$
3. The frequencies at which string is set into oscillations:$f_1=880Hzand{f}_2=1320Hz$
4. The range of the frequency: 500-1500Hz.

Understanding the concept of frequency

We can write the formula for frequency for n and n+1 harmonics. The difference between these two can give the fundamental frequency. By inserting the given values in this expression, we can get the values of the tension in the string.

Formula:

The velocity of the string in SHM,

$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{\tau }}{\mathbf{\mu }}}$ …(1)

The frequency of the string in motion,

$\mathit{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{\lambda }}$ …(2)

The resonant frequency of a body with n number of oscillations,

$\mathit{f}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}\mathbf{L}}\sqrt{\frac{\mathbf{\tau }}{\mathbf{\mu }}}$ …(3)

Calculations of the tension of the string

To find the tension in the string, the lower frequencyof the motion of the string is given using equation (1) as:

$f_1=\frac{n_1}{2L}\sqrt{\frac{\tau }{\mu }}$

The next higher frequency with n = n₁+ 1 is using same equation (3)is given as:

$\begin{array}{rcl}{f}_2& =& \frac{n_1+1}{2L}\sqrt{\frac{\tau }{\mu }}\\ & =& \frac{n_1}{2L}\sqrt{\frac{\tau }{\mu }}+\frac{1}{2L}\sqrt{\frac{\tau }{\mu }}\\ & =& f_1+\frac{1}{2L}\sqrt{\frac{\tau }{\mu }}\end{array}$

This gives the frequency difference of two oscillations which is given. Hence, the value of tension using this equation can be found as:

$\begin{array}{rcl}{f}_2-f_1& =& \frac{1}{2L}\sqrt{\frac{\tau }{\mu }}\\ \tau & =& 4L^2\mu {\left(f_2-f_1\right)}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 4{\left(0.300m\right)}^2\left(0.650\times {10}^{-3}kg/m\right){\left(1320Hz-880Hz\right)}^2\\ & =& 45.3N\end{array}$

Hence, the tension of the string is, 45.3N.