Answer to Question #126271 in Mechanics | Relativity for Che

Question

A rod, 1 cm2 x 200 cm and mass 2 kg, is clamped at its center. When vibrating longitudinally it emits its third overtone in unison with a tuning fork making 6000 vps. How much will the rod be elongated if, when clamped at one end, a stretching force of 980 N is applied to the other end?

Answer 1

Explanations & Calculations

• Velocity of the longitudinal waves through a dense medium = $\sqrt {\frac{E}{\rho}}$
• During the 3^rd overtone the length of the rod ($l$ )= $\frac{7\lambda}{2}$ when the rod is clamped at its center. therefore, the frequency of the third overtone

$\qquad\qquad \begin{aligned} \small f &= \small \frac{v}{\lambda}= \frac{7v}{2l} = \frac{7}{2l}\sqrt\frac{E}{\rho} \end{aligned}$

• $\small \rho = \large \frac{m}{volume } = \large\frac{2kg}{200\times10^{-6}m^3} = \small 10000kgm^{-3}$

• Since the rod is in resonance with the tuning fork, f =6000
• Therefore,

$\qquad\qquad \begin{aligned} \small 6000 & = \small \frac{7}{2\times 2m}\sqrt{\frac{E}{10000}}\\ \small E &= \small 1.17551\times 10^{11}Nm^{-2} \end{aligned}$

• Therefore, the elongation (e) could be calculated to be,

$\qquad\qquad \begin{aligned} \small \frac{F}{A} & =\small E \frac{e}{l}\\ \small e & = \small \frac{F l}{AE}\\ &= \small \frac{980N\times2m}{10^{-4}m^2\times (1.17551\times 10^{11}Nm^{-2})}\\ &= \small \bold{1.667\times 10^{-4}m}\\ &= \small \bold{0.1667mm} \end{aligned}$