(i) A wire having a linear mass density $6 \times 10^{-3}\ \text{kg m}^{-1}$ is stretched between two rigid…
(i) A wire having a linear mass density $6 \times 10^{-3}\ \text{kg m}^{-1}$ is stretched between two rigid supports with a tension of $540\ \text{N}$. The wire resonates at a frequency of $410\ \text{Hz}$. The next higher frequency at which the same wire resonates is $480\ \text{Hz}$. Find the length of the wire.
(ii) A $40\ \text{cm}$ wire having a mass of $3.2\ \text{g}$ is stretched between two fixed supports $40.05\ \text{cm}$ apart. In its fundamental mode, the wire vibrates at $220\ \text{Hz}$. If the area of cross-section of the wire is $10^{-6}\ \text{m}^2$, find its Young's modulus.
Solution
Sol. (i) Given, $\alpha = 6 \times 10^{-3}$ kg m$^{-1}$ and $T = 540$ N
Let frequency of $n$th harmonic be 410 Hz, then frequency of $(n + 1)$th harmonic be 480 Hz.
Fundamental frequency = $480 - 410 = 70$ Hz
$f_1 = \frac{1}{2l}\sqrt{\frac{T}{\alpha}} \Rightarrow 70 = \frac{1}{2l}\sqrt{\frac{540}{6 \times 10^{-3}}} = \frac{1}{2l} \times 300$
$\therefore$ Length of the wire, $l = \frac{150}{70} = 2.1$ m
(ii) Given, $L = 40$ cm, $M = 3.2$ g,
$\Delta L = 40.05 - 40 = 0.05$ cm,
$f = 220$ Hz and $A = 10^{-6}$ m$^{2}$
Young modulus, $Y = \frac{T/A}{\Delta L / L}$
Tension, $T = AY \frac{\Delta L}{L} = \frac{AY(0.05)}{40} = \frac{AY}{800}$
$\Rightarrow T = \frac{AY}{800} = \frac{10^{-6}Y}{800}$
$\alpha = \frac{M}{L} = \frac{3.2 \times 10^{-3}}{40 \times 10^{-2}} = 8 \times 10^{-3}$ kg m$^{-1}$
Frequency, $f = \frac{1}{2L}\sqrt{\frac{T}{\alpha}}$
or $220 = \frac{1}{2 \times 0.4} \sqrt{\frac{10^{-6} Y}{800 \times 8 \times 10^{-3}}}$
$176 = \sqrt{\frac{Y}{6.4 \times 10^{6}}}$
Young's modulus, $Y = 6.4 \times 10^{6} \times (176)^{2} = 1.98 \times 10^{11}$ Nm$^{-2}$
Answer: (i) Length $l = 2.1$ m
(ii) Young's modulus $Y = 1.98 \times 10^{11}$ Nm$^{-2}$