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# Answer to Question #126271 in Mechanics | Relativity for Che

Question #126271
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?
1
2020-07-14T08:51:41-0400

Explanations & Calculations

• Velocity of the longitudinal waves through a dense medium = "\\sqrt {\\frac{E}{\\rho}}"
• During the 3rd 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

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

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

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

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