91Ó°ÊÓ

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

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,

$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{τ}}{\mathbf{μ}}}$ …(1)

The frequency of the string in motion,

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

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

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

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

$f_1=\frac{n_1}{2L}\sqrt{\frac{\mathrm{τ}}{\mathrm{μ}}}$

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

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

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

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

Substitute all the value in the above equation.

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

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